Topic: Thermodynamics
A steel ball of mass 1 kg and specific heat 0.4 kJ/kg is at a temperature of 60 C. It is dropped into 1 kg water at 20 C. The final steady state temperature of water is
- A. 23.5 C
- B. 30 C
- C. 35 C
- D. 40 C
Answer: A. 23.5 C
Explanation: The heat lost by the steel ball equates to the heat gained by the water: [m * c * delta_t]_ball = [m * c * delta_t]_water. Substituting the values gives 1 * 0.4 * (60 - t) = 1 * 4.18 * (t - 20). This expands to 24 - 0.4t = 4.18t - 83.6. Rearranging terms yields 107.6 = 4.58t, resulting in t = 23.49 C, which is approximately 23.5 C.