✔ 最佳答案
Assume that there is no heat loss to the surroundings.
Heat transferred, q = m c ΔT
Density of water = 1.00 g/mL
Mass of the water = (50.0 mL) × (1.00 g/mL) = 50.00 g
Heat loss by the metal = Heat gained by the water
(100 g) × (0.240 J/g°C) × [(80.5 - T) °C] = (50.00 g) × (4.184 J/g°C) × (T - 21.5)
1932 - 24.0T = 209.2T - 4497.8
233.2T = 6429.8
Temperature at equilibrium, T = 27.6°C (to 3 sig. fig.)
Heat transferred from the metal to the water
= (100 g) × (0.240 J/g°C) × [(80.5 - 27.6) °C]
= 1270 J