Heat exchange

[victorias]

If two cups are in thermal contact with each other, one has 250 grams of water at 100 C and the other has 250 grams of water at 0C - I know that once thermal equilibrium is reached, the water in both cups will be at 50 C. This is just based on intuition because both cups have equal amount of water.

What if the two cups contained different amounts of water? For example, if cup 1 has 250 g and cup 2 has 50 g. I am confused as to how I can find out the temperature after thermal equlibrium is reached. What equation should we use here? I was trying q = mct

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[aldol16]

You're using the right equation. So the heat lost by one is gained by the other. You know the initial temperatures of each cup. Set up the equations and find the final temperature.

 

[victorias]

Here is my attempt:

delta q1 = delta q2
m1 c delta T1 = m2 c delta T2
(250 g) (1 cal/g * C) (4.2 J/cal) (T2 - 100 C) = (50 g) (1 cal/g * C) (4.2 J/cal) (T2 - 0 C)
T2 = 125

but this doesn't seem right because T2 should be different for cup 1 and cup 2
What am I doing wrong here?

Thanks

 

[poochimaster]

Hi Victorias,

What assumption leads you to believe that T2 should be different for cup 1 and cup 2? Isn't the definition of thermal equilibrium for two objects to have the thermal energy, and thus the same temperature?

Also, you may want to double check your calculations; intuitively it doesn't make sense for T2 to be greater than 100 C (since the hottest cup is 100 C and it is in contact with a colder cup).

 

[victorias]

I was thinking that since the 2 cups have different amounts of water in them, they would end up with different temperatures but with thermal equilibrium, it makes sense that they should end up with the same temperature.

Intuitively, I know that temperature of cup 1 should decrease and cup 2 should increase and since they both have different amounts of water, it would not be exactly in the midpoint at 50 C.

I redid my calculations (this time using a calculator to make sure, instead of rounding) but I am still getting 125

(250 g) (1 cal/g * C) (4.2 J/cal) (T2 - 100 C) = (50 g) (1 cal/g * C) (4.2 J/cal) (T2 - 0 C)
1050T2 - 105000 = 210T2 -0
840T2 = 105000
T2 = 105000/840
T2 = 125

 

[poochimaster]

Hey Victorias,

I found your problem; just a slight mistake in setting up the equation. The change in temperature in the hot cup should be 100 C - T2, not T2 - 100 C!

Rewritten, it should look like:
(250 g) (1 cal/g * C) (4.2 J/cal) (100 C - T2) = (50 g) (1 cal/g* C) (4.2 J/cal) (T2 - 0 C)

EDIT: this is wrong. See correction below.

 

[victorias]

Thanks

 

[poochimaster]

Sorry Victorias, I made a mistake with the correction above; completely ignore it. What you did with the set-up was actually correct, since delta T = Tf - Ti, and what I did violates that.

The mistake that you made was with the delta Q1 = delta Q2 set-up; it should actually be (-) delta Q1 = delta Q2, since heat is being lost in the first cup and heat is being gained in the second cup. With this correction you should get the answer of a final temperature in both cups being 83 C.

 

[aldol16]

Yes, the heat lost by the first cup, -q1, is equal to the heat gained by the second cup, +q2. Always remember your sign conventions and use your intuition to check if the answer is in the right range.