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Two blocks of the same metal are the same size but are at different temperatures, T1 and T2. These blocks of metal are brought together and allowed to the same temperature. Show that the entropy change is given by

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Complete question is missing, so i have attached it.

Answer:

ΔS = nC_p*In[(T1 + T2)²/4(T1•T2)]

Explanation:

Since the two blocks are kept in contact together, the final temperature will be given written as:

T_f = (T1 + T2)/2

Now, the entropy change of the system would be expressed as;

ΔS_sys = ∫(dq_rev)/T

This can be broken down into;

ΔS_sys = nC_p∫(1/T)dT

For the first block, when we integrate with boundaries of T_f and T1, we have;

ΔS1 = nC_p[In (T_f/T1)]

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Similarly, for the second block, integrating with boundaries of T_f and T2, we have;

ΔS2 = nC_p[In (T_f/T2)]

Thus, total change in entropy will be;

ΔS = ΔS1 + ΔS2 = nC_p[In (T_f/T1)] + nC_p[In (T_f/T2)]

This gives;

ΔS = nC_p*In[T_f²/(T1•T2)]

Earlier, we saw that T_f = (T1 + T2)/2

Thus;

ΔS = nC_p*In[(T1 + T2)²/4(T1•T2)]

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