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

Complete question is missing, so i have attached it.

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)]

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)]

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

it is the most commonly accepted one