A 0.10 kg piece of copper at an initial temperature of 95degC is dropped into 0.20 kg of water contained in a 0.28 kg aluminum calorimeter. The water and calorimeter are initially at 15degC. What is the final temperature of the system when it reaches equilibrium?

Question

(Cp of Copper=387J/kg * degC; Cpof Aluminum=899J/kg * degC; Cp of Water=4186J/kg * degC)

someone gave me a very long and complicated formula for this. does anyone know a simple way to solve this problem? thanks for any help

Answer 1

At thermal equilibrium, the heat lost by the hot copper will equal the heat gained by the water and calorimeter. By the time you have turned that fact into an equation for the final temperature, with symbols for the different masses and Cp's, you will have a lot of terms in a rather cumbersome equation. There is no other way to do it.

0.10*387*(95 - T) = 0.28*899*(T-15) + 0.20*4186*(T-15)

38.7 (95-T) = 1088.9(T-15)
188.9 T + 38.7 T = 16344 + 3676
227.6 T = 20020.
T = 88.0 C

No gurantees on the math.

Answer 2

this is what i got:

0.10*387*(95-T)=0.28*899*(T-15)+0.20*4186*(T-15)

38.7(95-T)=1088.92(T-15)

1088.92T+39.7T=16333+3676

1127.62T=20009.5

and i''m confused from here

Answer 3

You're right. I left out a zero in the 1088.9 T term. You changed 38.7 to 39.7, but that is not a big change.

1088.92T +38.7T = 16333+3676
1127.5 T = 20,009
T = 17.7 C. Not much of an increase. That is because the heat capacity of the small piece of copper is much less than that of the calorimeter and water.

Answer 4

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