Question

Calculate change in H and change in S when a 10.0 kg block of Fe at 500 K is placed in contact with a 5.0 kg block of Fe at 273 K. The specific heat capacity of Fe is 0.449 J K^-1-g^-1. Ignore the effect of the container.

Answer 1

If we kep the both Fe blocks in contact with each other heat will transer from block at higher temperature (500 K) to block at lower temperature (273 K) untill temparature of both the bolcks will be equal (386.5 K).

The change in temperature during heat transfer (T) = 113.5 K

During the process volume is constant therefore H = U = q

Quantity of heat transfered (q) = specific heat x mass x T

= 0.449 x 10 x 10 x 10³ J

= 44.9 x 10³ J = 44.9 kJ

Change in Enthalpy H = 44.9 kJ

Change in Entropy S = q/T

Change in entropy at temperature 386.5 K is

  S = 44.9/386.5 kJK^-1

= 116.2 JK^-1