Question

A 31.7 g wafer of pure gold initially at 69.7 ∘C is submerged into 64.1 g of water at 27.4 ∘C in an insulated container.

What is the final temperature of both substances at thermal equilibrium?

Answer 1

The heat lost by the gold will equal the heat gained by the water. Let Tf equal the final (equilibrium) temperature.

specific heat of water= 4.18, specific heat of gold= .128

Heat lost by gold = 31.7 g x 0.128 J/gC x (69.7 - Tf)
Heat gained by water = 64.1 g x 4.18 J/gC x (Tf - 27.4)

31.7 g x 0.128 J/gC x (69.7 - Tf) = -64.1 g x 4.18 J/gC x (Tf - 27.4)
4.0576 (Tf - 69.7) = - 267.938 (Tf - 27.4)

4.0576Tf - 282.815 = - 267.938 Tf + 7,341.5
4.0576Tf + 267.938 Tf = 7,341.5 + 282.815
271.9956 Tf = 7624.315
Tf = 28.03