Question

A 10.0 g piece of iron (C = 0.443 J/g ^oC) initially at 97.6 ^oC is placed in 50.0 g of water (C = 4.184 J/g ^oC) initially at 22.3 ^oC in an insulated container. The system is then allowed to come to thermal equilibrium. Assuming no heat flow to or from the surroundings, calculate

the final temperature of the metal and water

the change in entropy for the metal

the change in entropy for the water

the change in entropy for the universe

Answer 1

heat is calculated .... q = mcΔT
heat lost by gold = heat gained by water
-mcΔT (Fe) = mcΔT (water)
Let T_Fe = Ti of gold,

and T_w = Ti of water, Tf is the final temp of both

-mc(T_f – T_Fe) = mc(T_f - T_w)
-10g x 0.45 J/gC (Tf – 97.6^o C) = 50g x 4.18 J/gC (Tf – 22.3^o C)
-4.5Tf +439.2 = 209Tf – 4660.7
213.5Tf =5099.9
Tf = 5099.9 / 213.5
Tf = 23.89^O C