Two solid masses, A and B, each of 1 Kg, are placed in thermal contact while rem

Question

Two solid masses, A and B, each of 1 Kg, are placed in thermal contact while remaining adiabatically isolated from their surroundings. The initial temperatures are TA = 400 K and TB = 300 K, and the specic heats, in KJ Kg^1 K^1 , are CA = 1 and CB = 1 + 2 *10^-3 T .
(a) Calculate the nal common temperature, assuming that the process is one of heat transfer only.
(b) Calculate the change in the entropy of each body and the change in the entropy of the composite system. Is your result consistent with the process being irreversible? Why?

I understand that

so for A:

and for B:

But I have been unable to find part A because of the specific heat of B being equal to:

Explanation / Answer

a)Given that mass m A=mB = m = 1.0 kg Heat capacities CA = 1x 10^3J Kg^-1 K^-1 CB = (1 + 2 *10^-3 T) x 10^3 3J Kg^-1 K^-1 TA = 400 K and TB = 300 K --------------------------------------------------------------------------------- Let Tx be the final tmeperature. In the steady state, the energy trasferred through the other one material in a certain time must be equal to that trasferred through the one material in the same time. QA = QB mCA (TA - Tx) = mCB (Tx - TB) CA (TA - Tx) = CB (Tx - TB) (1x 10^3J Kg^-1 K^-1 )[ 400 K - Tx] = ((1 + 2 *10^-3 T) x 10^3 3J Kg^-1 K^-1)[ Tx - 300 K] (1)[ 400 K - Tx] = (1 + 2 *10^-3 T)[ Tx - 300 K] 1/ (1 + 2 *10^-3 T) = [ Tx - 300K]/(400 K - Tx) 0.998 = [ Tx - 300K]/(400 K - Tx) (0.998)[ 400 K - Tx] = Tx - 300 K Tx= 349.9499 K

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