Question

Calculate ΔH and ΔS 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

the ice block of mass 10 kg at 500k loses heat while 5 kg block at 273 k gains heat at 273 K gains heat

Both the blocks reach equilibrium temperature. let this be T

Heat lost by Fe block at 500 k = heat gained by Fe block at 273K

mass of Fe block at 500 K* specific heat of Fe block* (500-T) =    10*Cp*(500-T) = 5*Cp*(T-273)

Cp =specific heat of Fe block

2*(500-T)= T-273

3T= 1273 T = 1273/3=424 deg.c

delH =mass* specific heat* temperature difference= 10*0.449*(500-424)=341.24 Joules

for Fe block at 500K

entropy change= m*Cp ln (424/500)   =10*0.449*ln(424/500)=-0.74029 j/K

for Fe blcok at 273K entropy change = 5*0.449*ln (424/273)=0.988 J/L

total entropy change = 0.988-0.74029=0.247713 J/K