A 0.30-kg lead ball is heated to 90.0 °C and dropped into an ideal calorimeter containing 0.50 kg of water initially at 20.0 °C. What is the final equilibrium temperature of the lead ball? The specific heat capacity of lead is 128 J/(kg · C°); and the specific heat of water is 4186 J/(kg · C°).

Question

Quizplus · Accepted Answer

Using the principle of conservation of energy, we can equate the heat gained by the water and the calorimeter to the heat lost by the lead ball.

The amount of heat gained by the water and the calorimeter:
Q1 = mcΔT1 = (0.5 kg)(4186 J/(kg·°C))(Tf - 20.0 °C)

The amount of heat lost by the lead ball:
Q2 = mcΔT2 = (0.30 kg)(128 J/(kg·°C))(90.0 °C - Tf)

Since Q1 = Q2:
(0.5 kg)(4186 J/(kg·°C))(Tf - 20.0 °C) = (0.30 kg)(128 J/(kg·°C))(90.0 °C - Tf)

Solving for Tf:
Tf = 21.3 °C

Therefore, the final equilibrium temperature of the lead ball and water is 21.3 °C. The answer is (C).

A 030-Kg Lead Ball Is Heated to 90