A 79.1 g piece of metal at 90.0°C is placed in 127 g of water at 21.0°C contained in a calorimeter. The metal and water come to the same temperature at 25.3°C.

Question

A 79.1 g piece of metal at 90.0°C is placed in 127 g of water at 21.0°C contained in a calorimeter. The metal and water come to the same temperature at 25.3°C. How much heat (in J) did the metal give up to the water? (Assume the specific heat of water is 4.18 J/g·°C across the temperature range.)

Answer 1

heat lost by the metal = heat gained by the water

q = mC∆T

heat gained by water = (127 g)(4.18 J/g/deg)(25.3 - 21.0) = (127 g)(4.18 J/g/deg)(4.3 deg) = 2283 Joules

So, since heat gained by water must equal heat lost by metal, this (2283 J) is the heat lost by the metal.