Question

1. Two identical bars, each of mass m and specific heat c, are initially kept at temperatures T₀and 5T₀respectively. The bars are brought into contact until equilibrium is established. Assuming the entire experiment took place in thermal insulation from the surroundings; find the entropy change of the system.

Answer 1

mass of each bar = m

specific heat of each bar=c

Initial temperature of one bar = To

Initial temperature of second bar = 5To

let the final temperature of system after achieveing thermal equilibrium be T

heat gained by cold bar = heat lost by hot bar

heat gained by cold bar = Hg = m*c*(T-To)

heat lost by hot bar = Hl = m*c*(5To-T)

m*c*(T-To) = m*c*(5To-T)

T=3To

Change in entropy of a body of mass m and specific heat capacity c changing from temperature T1 to T2 is given by

Change in entropy of cold bar =

Change in entropy of hot bar =

Total chnage in entropy of system = m*c*ln(3)+m*c*ln(0.6) = m*c*[ln(3)+ln(0.6)] = m*c*ln(1.8) = 0.588*m*c

This is the required answer