Logic Questions

[Shadow Fury]

Keep your minds active! I'll be asking here a few questions regarding logic and see if you can answer them. They are in order of difficulty, meaning that the first question is the easiest and the last one is the hardest.

1- You have a glass of water at 20 degrees Celsius and some ice cubes at 0 degrees Celsius. You drop the ice cubes into the water and after a few minutes, the temperature of the water drops to 15 degrees Celsius. What is the temperature of the ice?

2- Three friends have a box with an amount of money. Each friend has to put twice the amount of money in the box then take $20 away. For example, if there are $50, the first friend will but $50 and take away $20, leaving $80 behind. Afterwards the second friend will repeat the procedure (add $80 then take away $20) and then the third friend will do the same. When they do this for real for the amount of money in the box, $0 are left behind. What was the starting amount of money?

3- You have 8 pills. 7 of them are safe to swallow but one of them is poisonous. All pills have the same weight except the poisonous one which is slightly heavier. You have a balance in front of you but you can only use it twice. How would you recognize the poisonous pill by using the balance?

 

[BloodDrunk]

I only figured out first.
0° Celsius, because if it were hotter than that, it wouldn't be ice anymore.

 

[Deleted member 238589]

1. I'm guessing the ice melted, and it combined with the rest of the water, so it's 15°.

2. 50$.

3. I think there are multiple ways to do it, but the easiest is to pick one random pill and separate it from the rest. Do that until you notice a difference in weight on the side where one pill is, since the other 7 have the same weight.

 

[Arad MNK]

1. The Ice would melt after that. So, 15°?

2. 17.5$, If happened three times.

3. We create 3 groups of Pills:
First with 3 pills,
Second with another 3 pills,
Third we pick the other 2 pills.

We check the First and second Group:
2 Conditions happen:
1. They are the Same weight
2. One goes down and one goes up

For 1: We check Group 3. Walla.

For 2: We pick the Heavier group. (Down)
We randomly choose one of the Pills, and Put it somewhere.
We pick the other two and use them on the balance.
2 Conditions happen:

1b. They have the same Weight
2b. One goes down and one goes up

For 1b: The one we picked is the Poisonous one

For 2b: It's obvious ha? The one goes down is the Poisonous one


Good Logical matter, Shadowfury!

 

[Shadow Fury]

Good job guys but unluckily nobody managed to answer all questions correctly. I won't reveal the answers yet, let's see others' responses.

By the way imperator, you can use the balance two times only. Your method would require several tries (removing pills and checking the weight is counted!)

 

[Arad MNK]

Uhm... I know It looks stupid, But is the (Melted Ice) 5°?

 

[Deleted member 238589]

...

3. Then you just place 4 on one side and 4 to the other. A heavier pile will contain a poisonous pill. That way you avoid the pile itself, without needing to pinpoint the poisonous pill.

 

[Shadow Fury]

...

3. Then you just place 4 on one side and 4 to the other. A heavier pile will contain a poisonous pill. That way you avoid the pile itself, without needing to pinpoint the poisonous pill.

But that's not what the question is asking. You must find out which one of the 8 pills is poisonous.

 

[kharza]

No, no you misunderstood me. 4-4. but I put the pills on the balance one by one, and I every time look on the balance. This way I know when I put the pill on it.

 

[Arad MNK]

Shadowfury, did you read my Way for the pills? I Believe it 100% works.

 

[Shadow Fury]

No, no you misunderstood me. 4-4. but I put the pills on the balance one by one, and I every time look on the balance. This way I know when I put the pill on it.

Still. Every time you check if a pill causes an unbalance. That's not allowed.

Arad MNK, I said I won't be revraling the answers yet!

 

[PublishedShadow]

1 - 0 degrees Celcius, otherwise it isn't ice.

2 - 10$ Because your first friend would place 10$ then take $20 leaving 0$. The other friends would place 0$ and since you can't take 20$ from 0$ the resulting dollar amount in the box is 0$.

3 - I would say you don't need the scale. If 7 are the same the poisonous one would look different.
However, If they are identical, (which you didn't specify above you just said they weighed almost the same) - the scale must be used, then you would have to: Place 4-4 taking the heavier four, then place 2-2 while again taking the heavier two, and then you would have to randomly pick between the two left since you have used your two tries.

 

[Arad MNK]

PublishedShadow, Randomness isn't allowed.

 

[PublishedShadow]

Ahem

Your answer is close unless they aren't the same weight, then you're left with 5 pills and one more weighing.
I'm still going with we are over thinking and that the poisonous pill looks different.

For 2: We pick the Heavier group. (Down)
We randomly choose one of the Pills, and Put it somewhere.
We pick the other two and use them on the balance.
2 Conditions happen:

EDIT
Nevermind I'm silly, that is the correct way.

 

[fladdermasken]

3- You have 8 pills. 7 of them are safe to swallow but one of them is poisonous. All pills have the same weight except the poisonous one which is slightly heavier. You have a balance in front of you but you can only use it twice. How would you recognize the poisonous pill by using the balance?

If you leave out two pills, and weigh two piles with three pills each, you have two different scenarios.

1. The piles weigh the same (first use of the scales). Ergo, they are all safe and the poisonous one is among the two you left out. Weigh these two to find out which one (second use of the scales).

2. One pile weighs more (first use of the scales). Ergo, the poisonous one is in the heavier pile. Take one pill out of this pile and weigh the other two (second use of the scales). If they weigh the same, the one you left out is poisonous. If they don't, the heavier one is poisonous.

Piece of cake.

 

[Shadow Fury]

I'm really about to die because of the laughter! Tomorrow I'll be saying the answers and why some answers are wrong (so funny that nobody scored 3/3)

 

[fladdermasken]

Just a reminder that if the answers you're going for are pure semamtics, you don't deserve to even utter the word "logic".

Otherwise the beauty of logic is that the answer confirms itself.

 

[Shadow Fury]

No semantics in the answers. I'm just surprised how people misinterpreted the questions or went for some random answers that really wipe out the *logic*.

 

[Dr Super Good]

1- You have a glass of water at 20 degrees Celsius and some ice cubes at 0 degrees Celsius. You drop the ice cubes into the water and after a few minutes, the temperature of the water drops to 15 degrees Celsius. What is the temperature of the ice?

Assuming a steady state has been reached then there is no ice (all ice melted) and the liquid which was the ice is now at 15 degrees C. Otherwise the ice would be at approximately 0 degrees C still (it might have impurities in it which raise or lower melting point or not be at sea level).

The logic is that water at normal atmospheric pressure in the form of ice has a melting point of 0 degrees. Due to a change in state from solid -> liquid a large amount of heat energy is required during which the temperature of the substance will not change. Due to heat transfer two contacting substances will converge towards the same temperature. As such if there is still ice (and its properties are still the same) then it will still be 0 degrees C since all heat it has absorbed has gone into changing state from solid -> liquid. If there is no ice and you are referring to the liquid which was once ice then that would be at 15 degrees C due to it converging towards the same temperature as the other water.

2- Three friends have a box with an amount of money. Each friend has to put twice the amount of money in the box then take $20 away. For example, if there are $50, the first friend will but $50 and take away $20, leaving $80 behind. Afterwards the second friend will repeat the procedure (add $80 then take away $20) and then the third friend will do the same. When they do this for real for the amount of money in the box, $0 are left behind. What was the starting amount of money?

Could be $0 since they can still try to take $20 away but get nothing instead (since the box is empty the transaction fails). Basically they add 0, try to take 20 from it (but fail so take nothing) and repeat 3 times.
0* 2 - 0 = 0
0* 2 - 0 = 0
0* 2 - 0 = 0

Otherwise if 20 has to always be removed (cannot fail) it would start with $17.50.
17.5 * 2 - 20 = 15
15 * 2 - 20 = 10
10 * 2 - 20 = 0
The logic behind this is that they have to add less than 20 (the amount they remove) so that it converges towards 0. If they add exactly 20 it will stay at a steady state forever. If they add more than 20 it will diverge towards infinity. One can rule out negative amounts since negative money does not exist in reality (has to be done with borrowing).

3- You have 8 pills. 7 of them are safe to swallow but one of them is poisonous. All pills have the same weight except the poisonous one which is slightly heavier. You have a balance in front of you but you can only use it twice. How would you recognize the poisonous pill by using the balance?

Assuming you mean simply the instruction to recognize then the poison pill will always displace (weigh more than when balanced) a single safe pill. Two safe pills (one on each side) will balance each other. This will not find you the poison pill but it will allow you to recognize it if you stumble across it.

Otherwise here is Arad MNK's solution re-explained.
Break 8 into 3:3:2 (A:B:C). Weigh A:B (balance use 1).
If A:B balance then break C into 1:1 (D:E). Weigh D:E (balance use 2) and heaviest one is a the poison pill.
If A heavier then break A into 1:1:1 (F:G:H). Weigh F:G (balance use 2). If F:G balance then H is poison pill otherwise heaviest one is poison pill.
If B heavier then break B into 1:1:1 (I:J:K). Weigh I:J (balance use 2). If I:J balance then K is poison pill otherwise heaviest one is poison pill.

Basically if any equal amount of pills are weighed and a balance occurs then the weighed pills do not contain the poison one (since there is only 1) so it has to be in any that were not weighed. Put in other words if the balance shows two equal sized piles as being equal weight then you can exclude their contents from the set of all pills as being safe. This allows you to perform a "tri-tree search" (probably given some stupid sounding name in science) in that a single test operation allows you to select one of 3 children nodes. Since this has a complexity of O(log3(N)) by breaking into even piles you can resolve 1 in 9 elements with only 2 tests. Since we only have 8 elements one of the nodes is incomplete (only 2 children instead of 3).

 

[Gilles]

Mother of God DSG. fladdy answered the third one; in english to boot!


Can we just hijack this thread into riddles? I like riddles.

 

[Xonok]

The first question is impossible to answer with the current information because:
1. Volume is not known.
2. Ice can have various temperatures, so it's not possible to assume.(to the people saying that ice is 0 degrees)

 

[Arad MNK]

If you leave out two pills, and weigh two piles with three pills each, you have two different scenarios.

1. The piles weigh the same (first use of the scales). Ergo, they are all safe and the poisonous one is among the two you left out. Weigh these two to find out which one (second use of the scales).

2. One pile weighs more (first use of the scales). Ergo, the poisonous one is in the heavier pile. Take one pill out of this pile and weigh the other two (second use of the scales). If they weigh the same, the one you left out is poisonous. If they don't, the heavier one is poisonous.

Piece of cake.

What's the difference of this with mine? (Ahem) NOTHING.

EDIT: DSG Too, all his answers were like me (With slight differences on 1.)

So, None get 3/3 yet right?

 

[Shadow Fury]

Ok here are the answers and why some answers are wrong:

1- The answer is 0 degrees C. The question clearly asks the temperature of ice so who cares if some melted, who cares about its volume? If its temperature is 0 and you put it in a hotter condition, it will stay at 0 until all ice melts (this has not been mentioned in the question) because the melting point of water is 0 and the temperature of ice does not change during this process. If you one tonne of ice at 0 degrees C it would be same thing of one gram of ice at 0 degrees C. I thought the majority would answer this correctly...

2- It's $17.50 (the calculations have been done above). Starting with $0 or $10 means that you are no more taking this question as one of logic but just a riddle. First of all, $0 is impossible because the question stated that there is an amount of money. $10 is just a shifty answer that is evidently not accepted.

3- Arad MNK got the right answer and flad + DSG simply rephrased him.
The question never mentioned that the poisonous pill looked different, otherwise what kind of dumb question would it be if you could recognize it by the naked eye?

DSG is the only who scored 3/3. Congrats to him!

 

[Xonok]

Ice typically isn't 0 degrees.

 

[Shadow Fury]

The version 2 of this thread will be opened when I'll manage to invent/find more logic questions. Currently I only have one and I plan to put at least 5 in the next version.

 

[Zwiebelchen]

Ice typically isn't 0 degrees.

Makers of such riddles usually don't care about physics.

If you were to explain to them that there is actually a way to have water in all three states (solid, liquid and gas) at the same time, their heads would probably explode.


Anyways, good luck in finding riddles that haven't been part of the professor Layton games already... ;D

 

[Flux]

Ice typically isn't 0 degrees.

Yes. Assume 1 atm.

@Zwiebelchen, Wow you played that game too. the pill question was there as far as i know.

 

[Dr Super Good]

because the melting point of water is 0

Only at a certain pressure and if it is pure water. The question really should have mentioned that the water involved melts at 0 since one cannot always make such an assumption.

The same can be said about boiling water. At sea level it boils at ~100 degrees Celsius but on top of Mount Everest it boils as low as 69 degrees odd. In a vacuum water boils well below normal sea level temperatures (20 degrees Celsius easily, probably even into the negatives).

 

[Gilles]

What goes up a chimney down, but not down a chimney up?

Please use spoiler tags of some form. And yes this is technically a logic question.

 

[Arad MNK]

What goes up a chimney down, but not down a chimney up?

Please use spoiler tags of some form. And yes this is technically a logic question.

Spoiler

The Big Bad Wolf!

 

[InfernalTater]

Keep your minds active! I'll be asking here a few questions regarding logic and see if you can answer them. They are in order of difficulty, meaning that the first question is the easiest and the last one is the hardest.

I hope you don't think these are clever.

1- You have a glass of water at 20 degrees Celsius and some ice cubes at 0 degrees Celsius. You drop the ice cubes into the water and after a few minutes, the temperature of the water drops to 15 degrees Celsius. What is the temperature of the ice?

Since I don't know the mass of the water or ice, and don't know the specific length of time of the "few minutes", I can't physically calculate whether or not the ice would still be ice after, or be entirely melted. If the ice is still frozen, then the remaining ice would have to still be 0 degrees Celcius in order to be frozen; if the ice is melted, it would theoretically be 15 degrees celcius as it mixes in with the water. However, the temperature does not balance instantly, and it is still possible that if the last of the ice has only just melted, that the temperature of the water has not yet been distributed evenly and the water from the melted ice, for a very short moment, will be cooler than the rest of the water.

Essentially, there are three potentially right answers to this question depending on the specific values of what is currently unknown.

2- Three friends have a box with an amount of money. Each friend has to put twice the amount of money in the box then take $20 away. For example, if there are $50, the first friend will but $50 and take away $20, leaving $80 behind.

You phrased this incorrectly. In order for there to be $50 in the box, and a friend add $50 into the box, he would have to be adding the amount of money in the box; not adding twice the amount of money in the box. I'll assume that you meant he would have to double the amount of money in the box (setting it to 200% of the previous value), not add double the amount of money in the box (which would set it to 300% of the previous value).

2- Three friends have a box with an amount of money. Each friend has to put twice the amount of money in the box then take $20 away. For example, if there are $50, the first friend will but $50 and take away $20, leaving $80 behind. Afterwards the second friend will repeat the procedure (add $80 then take away $20) and then the third friend will do the same. When they do this for real for the amount of money in the box, $0 are left behind. What was the starting amount of money?

Assuming that your math was correct but your phrasing was wrong, and you meant that they add the current amount of money to the box, not adding twice the amount of money in the box, the starting amount of money was $10, in which case he would add $10, and remove $20, resulting in a balance of $0. Assuming that your phrasing of adding twice the amount of money was wrong but your phrasing was right, and he was indeed adding twice the amount of money in the box, to the box, then the starting money would have to be $6.66 (and 0.66 cents), in which case he would add $13.34 before removing $20 from the box, again resulting in $0. Anything in between $0 and $6.66 would result in a remaining balance of $0, as, after adding twice the amount, there would be nothing left after attempting to remove $20.

3- You have 8 pills. 7 of them are safe to swallow but one of them is poisonous. All pills have the same weight except the poisonous one which is slightly heavier. You have a balance in front of you but you can only use it twice. How would you recognize the poisonous pill by using the balance?

I didn't think about this one long enough to come up with the right answer. I'll consider myself wrong.

If you leave out two pills, and weigh two piles with three pills each, you have two different scenarios.

1. The piles weigh the same (first use of the scales). Ergo, they are all safe and the poisonous one is among the two you left out. Weigh these two to find out which one (second use of the scales).

2. One pile weighs more (first use of the scales). Ergo, the poisonous one is in the heavier pile. Take one pill out of this pile and weigh the other two (second use of the scales). If they weigh the same, the one you left out is poisonous. If they don't, the heavier one is poisonous.

Piece of cake.

I hadn't thought of that.

 

[Shadow Fury]

Why are you still discussing this guys? The first question was meant to be a simple question of logic and not complicated science debates. Ice at 0 degrees will stay at 0 degrees during the melting phase (you had to understand that by yourself). Room temperature and atmospheric pressure were blatantly normal + the ice was pure for sure (it even rhymes...). If I had revealed more information, then not only wouldn't it be a question of logic but also too easy to respond.

As for the second one, it's not wrong or anything. The statement is clear; double the amount of money in the box. According to what you're saying, Tater, I'd have to say "Add one time the amount of money to the box". Doesn't make sense...

I'll request this thread to be closed before you start endless unneeded debates on what were meant to be simple questions of logic.

 

[InfernalTater]

Why are you still discussing this?

This is my first post in this thread.

The first question was meant to be a simple question of logic and not complicated science debates. Ice at 0 degrees will stay at 0 degrees during the melting phase (you had to understand that by yourself). If I had revealed more information, then not only wouldn't it be a question of logic but also too easy to respond.

You have to know the science behind it to determine whether or not the ice will still be ice in the first place. Depending on various factors, it may be completely melted by then and be a different temperature.

As for the second one, it's not wrong or anything. The statement is clear; add two times the amount of money in the box. According to what you're saying, Tater, I'd have to say "Add one time the amount of money to the box". Doesn't make sense...

2- Three friends have a box with an amount of money. Each friend has to put twice the amount of money in the box then take $20 away. For example, if there are $50, the first friend will but $50 and take away $20, leaving $80 behind.

Do you know math? You claimed that it started with $50, that he added $50, took away $20, and left a total of $80. This means that he added the amount of money that was already in the box ($50), not twice the amount of money that was in the box. If he added twice the amount, and then removed $20, it would have left $130. (50 + 2(50) - 20) = 130.

I'll request this thread to be closed before you start endless unneeded debates on what were meant to be simple questions of logic.

I'm still speaking on-topic, I'm just questioning your logic instead of ours.

You make too many threads like this and you always react poorly when we point out what is wrong.

Pretending you know what you're talking about, when you don't have the slightest clue, is the reason you don't know what you're talking about. If you would admit that you know nothing, and start working towards learning what you do not know, then you wouldn't have to pretend that you are wise anymore; you would actually be able to be wise.

 

[Shadow Fury]

I'm not saying you're speaking off-topic, I"m just saying it's useless to discuss this. People answered the questions and understood the second one correctly so I don't see the reason of the argument from the principle.

The only bad thing is that when I open thread, you try to find flaws in it rather than enjoying it. If you'd just keep seriousness away for a moment and try to have some fun, we wouldn't have such discussions, trust me. (fun in this thread maybe not but in Warcraft Memes yes!)

 

[InfernalTater]

I'm not saying you're speaking off-topic, I"m just saying it's useless to discuss this. People answered the questions and understood the second one correctly so I don't see the reason of the argument from the principle.

The only bad thing is that when I open thread, you try to find flaws in it rather than enjoying it. If you'd just keep seriousness away for a moment and try to have some fun, we wouldn't have such discussions, trust me. (fun in this thread maybe not but in Warcraft Memes yes!)

If you'd stop screwing up your posts in ways that are detrimental to the purpose of them (like failing logically in a thread about logic), I wouldn't have to point out what's wrong with them.

 

[Ash]

I hope you don't think these are clever.

Jerk.

Since I don't know the mass of the water or ice, and don't know the specific length of time of the "few minutes", I can't physically calculate whether or not the ice would still be ice after, or be entirely melted. If the ice is still frozen, then the remaining ice would have to still be 0 degrees Celcius in order to be frozen; if the ice is melted, it would theoretically be 15 degrees celcius as it mixes in with the water. However, the temperature does not balance instantly, and it is still possible that if the last of the ice has only just melted, that the temperature of the water has not yet been distributed evenly and the water from the melted ice, for a very short moment, will be cooler than the rest of the water.

There is one correct answer to this: ice is 0 degrees when it is ice. If it is still frozen, it is ice. If it is defrosted, it is water. This is different to surface temperature. There are many extra factors, but so long as it is ice, so long as it is in standard conditions, it will melt at a steady 0 degrees Celsius. Your next assumption that the water would be 15 degrees if the ice had totally melted is as wrong as it is stupid: that is very. Allow me to elaborate.

First, we need to make some assumptions as this is not what the question is about. The room temperature is 20*c (and so, water temperature is also initially 20*c), we have 100g of water, enough to quench a small thirst, and 20g of ice. The drink is to be stored in an insulted cup.

The final temperature will be 3.3*c. Why? Express it as an equation:

Wi×Hf + Wi×Sh×(Tf - 0*c) = Ww×Sh×(Twi - Tf) 
Where
Wi = weight of ice = 20g
Hf = Heat of fusion = 80 calories/gram
Ww = weight of water = 100g
Tf = final temperature,
Twi = initial water temperature = 20*c
Sh = Specific Heat for water = 1.0 calories per gram per *c

Substitute your values:

20×80 + 20×1×(Tf - 0*c) = 100×1×(20*c - Tf)
1600 + 20Tf = 2000 - 100Tf
120Tf = 400
Tf = 400/120 = 3.3*c

If you're going to post a pretentious and arrogant answer, at least check to make sure you're right. Which brings me nicely to the only reason I'm actually posting in this thread: stop being so pretentious and arrogant. You've been warned about this before. There are much better, much nicer ways to disagree with someone, and the way you're choosing to do so will not be tolerated at THW.

 

[InfernalTater]

Jerk.



There is one correct answer to this: ice is 0 degrees when it is ice. If it is still frozen, it is ice. If it is defrosted, it is water. This is different to surface temperature. There are many extra factors, but so long as it is ice, so long as it is in standard conditions, it will melt at a steady 0 degrees Celsius. Your next assumption that the water would be 15 degrees if the ice had totally melted is as wrong as it is stupid: that is very. Allow me to elaborate.

First, we need to make some assumptions as this is not what the question is about. The room temperature is 20*c (and so, water temperature is also initially 20*c), we have 100g of water, enough to quench a small thirst, and 20g of ice. The drink is to be stored in an insulted cup.

The final temperature will be 3.3*c. Why? Express it as an equation:

Wi×Hf + Wi×Sh×(Tf - 0*c) = Ww×Sh×(Twi - Tf) 
Where
Wi = weight of ice = 20g
Hf = Heat of fusion = 80 calories/gram
Ww = weight of water = 100g
Tf = final temperature,
Twi = initial water temperature = 20*c
Sh = Specific Heat for water = 1.0 calories per gram per *c

Substitute your values:

20×80 + 20×1×(Tf - 0*c) = 100×1×(20*c - Tf)
1600 + 20Tf = 2000 - 100Tf
120Tf = 400
Tf = 400/120 = 3.3*c

If you're going to post a pretentious and arrogant answer, at least check to make sure you're right. Which brings me nicely to the only reason I'm actually posting in this thread: stop being so pretentious and arrogant. You've been warned about this before. There are much better, much nicer ways to disagree with someone, and the way you're choosing to do so will not be tolerated at THW.

You are assuming that the mass of the water is 100g and that the mass of the ice is 20g. He did not provide values for these; the water could be any amount, and the ice could be any amount. I did not calculate the resulting temperature of 15 degrees celsius; it was stated in the original post that the resulting temperature of the water was 15 degrees celcius. While I do indeed believe that 15 degrees is quite a high temperature to result, and that realistically, ice would cool the water much more than a mere 5 degrees, I did not provide the temperatures; he did. Any logical flaw in the temperature is yet another flaw of his own post, not mine. Thus, if the ice is still frozen, it is 0 degrees celsius, or if it has melted, it is the same temperature as the water (15 degrees celsius), as it is mixed within it.

3.3 degree water won't just sit inside of 15 degree water, hanging out like neighbors. Your entire calculation is a wasted effort as you failed logically.

Go ahead, continue insulting my intelligence while providing incorrect equations. You only make yourself look ignorant.

There are much better, much nicer ways to disagree with someone, and the way you're choosing to do so will not be tolerated at THW.

I've seen much harsher ways of disagreeing with people from moderators including yourself.

I am not being terribly offensive; I did not even call him any names, as you have done to me. All I've done is pointed out that because his questions were not stated properly, that they couldn't be accurately answered, in attempt to persuade him to make his future posts more logically sound. He doesn't have to use logic, and he can continue posting the same way that he already does, but the same freedom that applies to him should apply to me in constructively criticizing his post. Disagreeing with moderators in off-topic threads about irrelvant subjects is not a violation of the rules.

If you want to make it against the rules to have a different opinion than moderators, feel free to petition that Ralle do so; until then, you'll just have to get used to interpreting THW's rules as liberally as possible to find an excuse to negrep or ban me until you feel better.

Yes, I am speaking off-topic now, but it is in response to you, a moderator, addressing me directly in this thread. If you didn't want me to respond here, you could have PM'd me, and then I would respond there; but since you chose to argue here, I respond here.

My original post, however, was on topic.

 

[Shadow Fury]

The ice did not melt fully, in fact, that's why I asked "the temperature of the ice". If there was no ice, then it would be illogical asking its temperature. I just wanted to trick some people and let them say 5 degrees C (and I actually managed to do that ^^).

 

[Ash]

He did not provide values as to the amounts of ice and water, and thus all of your calculations are null because you're using masses that you assumed. If there were 100 grams of water and 50 grams of ice, 20 grams of ice, or 1 gram of ice, the results are going to be quite different.

He did, however, state that the temperature of the water at the end was 15 degrees celsius. If the ice is completely melted and mixed in with the rest of the water, and the water is 15 degrees celcius, then the ice which has melted into water will be 15 degrees celcius as well. Granted, in order for this to be the case, the amount of ice would have to be a very tiny amount, or else it would cool the water much more significantly, however I didn't make up the resulting temperature of 15 degrees; the original post did.

And as I already stated before, if the ice is not melted, then obviously it is still 0 degrees celcius.

Thank you for being much more civil.

If we had had more ice or less water, there would not be enough energy in the water to melt all the ice. Once the temperature of the water hits zero, no more ice can melt.
I guess an interesting question is how much ice could be melted by 100 g of water at 20*c? Using the equation above, where the final temperature is zero and the weight of the ice is unknown, we have:
Wi×80 = 100×1×20

The maximum amount of ice that could be melted is:
Wi = 2000/80 = 25 grams
If the amount of ice exceeds 25 grams, the remainder will just float in 125 grams of water at zero.

Of course the calculations aren't null, they were made with assumptions as we don't have all the data. This is a forum, not a lab, so we're limited as to how far we can actually go with things. ShadowFury didn't mention that the cup was insulated, either.

 

[InfernalTater]

Thank you for being much more civil.

If we had had more ice or less water, there would not be enough energy in the water to melt all the ice. Once the temperature of the water hits zero, no more ice can melt.
I guess an interesting question is how much ice could be melted by 100 g of water at 20*c? Using the equation above, where the final temperature is zero and the weight of the ice is unknown, we have:
Wi×80 = 100×1×20

The maximum amount of ice that could be melted is:
Wi = 2000/80 = 25 grams
If the amount of ice exceeds 25 grams, the remainder will just float in 125 grams of water at zero.

Of course the calculations aren't null, they were made with assumptions as we don't have all the data. This is a forum, not a lab, so we're limited as to how far we can actually go with things. ShadowFury didn't mention that the cup was insulated, either.

He already stated that the resulting temperature of the water was 15 degrees celsius. Your calculations resulted in a temperature of 3.3 degrees celsius. Thus your entire equation was wrong because you clearly used more ice. The ice can only be 0 degrees celcius (as a solid), or the same temperature as the water that it is mixed in if it has melted entirely (aside from during the split second it takes for the melted ice to distribute throughout the cup). Since he stated that the temperature of the water at the end was 15 degrees, then ice which was completely melted into water would also have to be 15 degrees. Water doesn't just hang out with 3.3 degree and 15 degree bodies of water on top of each other like neighbors; it mixes evenly until it is all the same temperature. Thus, if any of the water in the cup is 15 degrees, all of the water in the cup will be relatively close to 15 degrees (external factors can affect the temperature of the water in different parts of the cup, however in an experiment all external factors should be eliminated).

 

[Ash]

He already stated that the resulting temperature of the water was 15 degrees celsius. Your calculations resulted in a temperature of 3.3 degrees celsius. Thus your entire equation was wrong because you clearly used more ice. The ice can only be 0 degrees celcius (as a solid), or the same temperature as the water that it is mixed in if it has melted entirely (aside from during the split second it takes for the melted ice to distribute throughout the cup). Since he stated that the temperature of the water at the end was 15 degrees, then ice which was completely melted into water would also have to be 15 degrees. Water doesn't just hang out with 3.3 degree and 15 degree bodies of water on top of each other like neighbors; it mixes evenly until it is all the same temperature. Thus, if any of the water in the cup is 15 degrees, all of the water in the cup will be relatively close to 15 degrees (external factors can affect the temperature of the water in different parts of the cup, however in an experiment all external factors should be eliminated).

I think what you're not getting is that as soon as it melts, it is no longer ice. Ice is 0*c. It is always zero. It's surface temperature may be different, but the ice is still zero. As soon as it is not zero, it is no longer ice. Try and think of it in binary terms.

 

[Xonok]

So you're saying glaciers aren't ice, because they're colder than 0.

 

[Shadow Fury]

I think he meant that if it is above 0 it is not ice even if he expressed himself with the wrong words. At and below 0 it's ice, that's it.

 

[InfernalTater]

I think what you're not getting is that as soon as it melts, it is no longer ice. Ice is 0*c. It is always zero. It's surface temperature may be different, but the ice is still zero. As soon as it is not zero, it is no longer ice. Try and think of it in binary terms.

That's an argument of semantics.

If you crushed a rock into powder, you could still refer to the powder as a rock, having known that it used to be a rock, despite knowing that it is no longer a rock. In its immediate state it is no longer a rock, however it could still be transformed back into a rock with heat and pressure. Somebody who never saw it as a rock would describe it as powder, however the owner of the rock, whose rock has been crushed, would refer to it as a ground up rock, just as a person whose ice has been melted (if the ice were of significant enough importance for its original form to be worth noting) may refer to the water as "melted ice" rather than simply calling it water, in order to specifically address that it wasn't always water.

Since his question was attempting to avoid specifics so as to not hint at the answer, it would not surprise me if it stated "ice" for the sake of not revealing the resulting form of the ice.

 

[Ash]

That's an argument of semantics.

If you crushed a rock into powder, you could still refer to the powder as a rock, having known that it used to be a rock, despite knowing that it is no longer a rock. In its immediate state it is no longer a rock, however it could still be transformed back into a rock with heat and pressure. Somebody who never saw it as a rock would describe it as powder, however the owner of the rock, whose rock has been crushed, would refer to it as a ground up rock, just as a person whose ice has been melted (if the ice were of significant enough importance for its original form to be worth noting) may refer to the water as "melted ice" rather than simply calling it water, in order to specifically address that it wasn't always water.

Since his question was attempting to avoid specifics so as to not hint at the answer, it would not surprise me if it stated "ice" for the sake of not revealing the resulting form of the ice.

Semantics is an important thing. If the Duke of Wellington and the Earl of Sandwich swapped places, we'd be walking on sandwiches and eating wellingtons. I'd say your example depends on the granularity. His question didn't avoid these specifics, it was just logic: is it ice? It's can't be any higher than zero.

 

[BloodDrunk]

Just stop this crap and give us more Logic Question!

 

[Shadow Fury]

BloodDrunk is right. Ok, here is a question NOT regarding science...

You have a grill and 3 steaks. Each of the two sides of a steak requires 2 minutes to be cooked. The grill can only sustain 2 steaks at a time. What is the minimum time required to cook all steaks and how would you do that?

 

[GhostThruster]

lets call 'em steaks A, B, and C

first, slap on A & B on side 1. that's 2 minutes

then, turn over A to side 2 and swap B with C. that's 2 more minutes

finally, remove A (which has been cooked on both sides), turn over C, and return B on side 2. that's a total of 6 minutes

 

[BloodDrunk]

lets call 'em steaks A, B, and C

first, slap on A & B on side 1. that's 2 minutes

then, turn over A to side 2 and swap B with C. that's 2 more minutes

finally, remove A (which has been cooked on both sides), turn over C, and return B on side 2. that's a total of 6 minutes

Damn you!
I had exactly the same idea but I couldn't not read yours haha =P

 

[Shadow Fury]

Correct GhostThruster!
Well, v2 of this thread will have to attend... but if you want to share logic questions here, do that! I wanna solve some as well