Impact of container on final temperature of system

[sharpnova]

Homework Statement

A container contains a given mass of ice and a given mass of water both at 0 degrees Celsius. A given mass of aluminum is heated up to some temperature and submerged in the water-ice mixture. No heat enters or leaves the container. Given the specific heat of water and Aluminum and the energy it takes to melt a kilogram of ice, find the final temperature of the solution.

If we let the container absorb some heat and it has a given mass and a given specific heat, what will be the final temperature of the solution.

Homework Equations

Q = C*M*T_change

The Attempt at a Solution

I have solved the first part of the question. Just two equations in the unknown (final temperature) then solved for it.

The second part has me stumped though. It seems like I'd need to know the initial temperature of the container to solve this one. But no temperature was given for the container. Am I misunderstanding the question or overlooking something subtle? Or should I just assume the container starts at 0 degrees Celsius?

 

[SteamKing]

Homework Statement

A container contains a given mass of ice and a given mass of water both at 0 degrees Celsius. A given mass of aluminum is heated up to some temperature and submerged in the water-ice mixture. No heat enters or leaves the container. Given the specific heat of water and Aluminum and the energy it takes to melt a kilogram of ice, find the final temperature of the solution.

If we let the container absorb some heat and it has a given mass and a given specific heat, what will be the final temperature of the solution.

Homework Equations

Q = C*M*T_change

The Attempt at a Solution

I have solved the first part of the question. Just two equations in the unknown (final temperature) then solved for it.

The second part has me stumped though. It seems like I'd need to know the initial temperature of the container to solve this one. But no temperature was given for the container. Am I misunderstanding the question or overlooking something subtle? Or should I just assume the container starts at 0 degrees Celsius?

Remember, no heat enters or leaves the container, according to the problem statement.

What does this tell you about the temperature of the container and the water-ice mixture?

 

[sharpnova]

I'm guessing that means it starts at 0 degrees Celsius but I don't understand why. I was thinking the "no heat enters or leaves" was just a magical constraint because if the container was cooler than room temperature then it should be bringing heat into the system... or if it's warmer then it should be radiating heat

 

[SteamKing]

I'm guessing that means it starts at 0 degrees Celsius but I don't understand why. I was thinking the "no heat enters or leaves" was just a magical constraint because if the container was cooler than room temperature then it should be bringing heat into the system... or if it's cooler then it should be radiating heat

Are you studying physics or magic?

Heat cannot be transferred to or from something without that something being at a different temperature than its surroundings. This is one of the laws of thermodynamics, specifically, the Zeroth Law:

https://en.wikipedia.org/wiki/Laws_of_thermodynamics

 

[sharpnova]

Sorry I wasn't clear. When I said that "I was thinking the 'no heat enters or leaves' was just a magical constraint" I was saying that I thought that a somewhat nonsensical constraint had been placed on the system in order to simplify the problem.

I assume by emphasizing that the constraint was not magical, that you are saying that it would actually be exhibited in this experiment.

So are you saying that the container would be at room temperature at the start?

Then I would think I'd need to know the room temperature to solve this problem.

If the container started at 0 degrees Celsius then it would be slightly cooling the room and the room would be slightly heating the container.

Or are you saying that the room-temperature and container temperature are both starting at 0?

Because even in this case the heated metal would raise the temperature of the container/water system which would in turn slightly cool the room and lose some of the energy that would be going into the container to the room.

 

[sharpnova]

Anyone?

 

[Chestermiller]

Anyone?

Suppose that the container is capable of exchanging heat with its contents, but is insulated on the outside, so it can't exchange heat with the room. Also assume that the container is initially at 0 C. Also assume that not all the ice melts when the system equilibrates with the aluminum. What effect do you think including the container vs not including the container would have?

 

[sharpnova]

I believe it would then reduce the final temperature of the system since it would have a resistance to temperature increase that would suck up some of the thermal energy (and started at a lower temperature than the system's average at the start)

Assuming I have that right I think I understand it now.

I wasn't sure if I should assume the container started at 0 inside and didn't exchange with outside

Thanks for your help (if I'm understanding you correctly)

 

[Chestermiller]

I believe it would then reduce the final temperature of the system since it would have a resistance to temperature increase that would suck up some of the thermal energy (and started at a lower temperature than the system's average at the start)

Assuming I have that right I think I understand it now.

I wasn't sure if I should assume the container started at 0 inside and didn't exchange with outside

Thanks for your help (if I'm understanding you correctly)

If all the ice doesn't melt, the final temperature of the contents must stay at 0C. The container temperature can't rise above 0 C because, if it did, the contents would cool it back down. So the container would stay at 0C, and its change in internal energy would be zero. So as long as there is ice remaining in the container, the container has no effect on the final state of the system. However, if all the ice melts, the container can increase in temperature along with the contents, and then, it would have an effect.