one kilogram of iron at 80ºC is dropped into 0.5L of water at 20ºC. Given that the specific heat of water is 1
one kilogram of iron at 80ºC is dropped into 0.5L of water at 20ºC. Given that the specific heat of water is 1 cal/g.K and that of iron is 0.107 cal/g.K, calculate the final equilibrium temperature of the system and the increase in entropy?
More Questions From This User See All
Answers & Comments
Verified answer
Let T be the final equilibrium temp. and every thing is in a totally insulated system
Then 1*0.107*(80-T) = 0.5*1*(T-20)
8.56-0.107*T = 0.5*T-10
0.607*T=18.56
T=18.56/0.607=30.57 C
Cp iron=0.107*4.1858 =0.448
entropy change = 0.448*ln(T2/T1)
=0.448*ln(353/303.57)=0.0676 J/K