Question:
A 2 kg block of copper is initially at a temperature of 100°C. The block is placed in contact with a large reservoir of water at 10°C. The block and the water reach a final equilibrium temperature of 20°C. Calculate the heat transferred to the water from the block.
Given: Specific heat capacity of water (c_w) = 4186 J/kg°C Specific heat capacity of copper (c_c) = 390 J/kg°C Change in temperature of the block (ΔT_c) = 100°C - 20°C = 80°C Change in temperature of the water (ΔT_w) = 20°C - 10°C = 10°C
Useful formula: Q = mcΔT
Answer:
First, let's calculate the heat lost by the copper block and the heat gained by the water.
Heat lost by copper block (Q_c): Q_c = mcΔT_c Q_c = (2 kg)(390 J/kg°C)(80°C) Q_c = 62400 J
Heat gained by water (Q_w): Q_w = mcΔT_w Q_w = (2 kg)(4186 J/kg°C)(10°C) Q_w = 83720 J
The heat transferred to the water from the block can be calculated by subtracting the heat lost by the block from the heat gained by the water:
Heat transferred = Q_w - Q_c Heat transferred = 83720 J - 62400 J Heat transferred = 21320 J
Therefore, the heat transferred to the water from the block is 21320 J.