A 155.0 g piece of copper at 168 oc is dropped into 250.0 g of water at 20.9 oc. (the specific heat of copper is 0.385 j/goc.) calculate the final temperature of the mixture. (assume no heat loss to the surroundings.)

Note that

The specific heat of water is 4.184 J / (g - C)

Let x = the final equilibrium temperature.

Heat loss by the copper is

(155 g) * (0.385 J / (g - C)) * (168 - x C) = 59.675 (168 - x) J

Heat gain by the water is

(250 g) * (4.184 J / (g-C)) * (x - 20.9 C) = 1046 (x - 20.9) J

Because there is no heat loss to the surroundings, therefore

59.675 (168 - x) = 1046 (x - 20.9)

168 - x = 17.5283 (x - 20.9)

18.5283x = 534.3415

x = 28.84 °C

Answer: 28.8 °°C nearest tenth)