Question

9. Overall change in entropy. A copper penny, initially at temperature Ti, is placed in contact with a large block of ice that serves as a heat reservoir and has a constant temperature Tres (well below freezing). Take the penny's heat capacity to have the constant value C, and specify Ti 7 Tres (by a finite amount). The following questions pertain after the joint system has come to thermal equilibrium. (a) What are the entropy changes of the penny and of the ice block? (b) What sign does the total change in entropy have [according to your calculations in part (a)]? (c) Is the sign independent of whether the penny was hotter or colder than the ice block?

Answer 1

(a) final temperature of copper penny = T_res

Heat transfer to copper penny = C×∆T = C×(T_res - T_i)

Entropy change of the ice block = - C(T_res - T_i)/T_res

Entropy change of copper penny = $\int$C/T dT = C.ln(T_res/T_i)

Total entropy change , ∆S_ice + ∆S_penny = C( T_i -T_res)/T_res + C.ln(T_res/T_i)

= C.T_i/T_res - C + C.ln(T_res/T_i)

= C ( T_i/T_res - ln(T_i/T_res) -1 )

(b) the total change in entropy should have positive sign .

(c) whenever a hot body comes in contact with cold body, heat flow would takes place from hot to cold. And entropy change of system+ surroundings would be positive .

If penny is hotter then heat flow from penny to ice otherwise reverse order .

The sign is independent of whether the penny was hotter or colder than the ice block .