A 31.1 g wafer of pure gold, initially at 69.3 _c, is submerged into 64.2 g of water at 27.8 _c in an insulated container. what is the final temperature of both substances at thermal equilibrium?

Given:

Ma = 31.1 g, the mass of gold

Ta = 69.3 °C, the initial temperature of gold

Mw = 64.2 g, the mass of water

Tw = 27.8 °C, the initial temperature of water

Because the container is insulated, no heat is lost to the surroundings.

Let T °C be the final temperature.

From tables, obtain

Ca = 0.129 J / (g-°C), the specific heat of gold

Cw = 4.18 J / (g-°C), the specific heat of water

At equilibrium, heat lost by the gold - heat gained by the water.

Heat lost by the gold is

Qa = Ma*Ca * (T - Ta)

= (31.1 g) * (0.129 J / (g-°C) ( * (69.3 - T °C) -

= 4.0119 (69.3 - T) j

Heat gained by the water is

Qw = Mw*Cw * (T-Tw)

= (64.2 g) * (4.18 J / (g-°C)) * (T - 27.8 °C)

= 268.356 (T - 27.8)

Equate Qa and Qw.

268.356 (T - 27.8) = 4.0119 (69.3 - T)

272.3679T = 7738.32

T = 28.41 °C

Answer: 28.4 °C

The final temperature of both substances at thermal equilibrium is 28.4 degrees Celsius.