Question

A 250 g piece of brass at 95C is placed into 500g of water at 25C in an insulated container. calculate the final equlilbtium temperature (heat) of the brass in the insulated container

Answer 1

Use conservation of energy (no heat lost to surroundings):

m_brass c_brass (T_brass,i − T_f) = m_water c_water (T_f − T_water,i)

Take c_brass ≈ 0.380 J/g·°C and c_water = 4.18 J/g·°C.

m_brass = 250 g, T_brass,i = 95 °C
m_water = 500 g, T_water,i = 25 °C

250(0.380)(95 − T_f) = 500(4.18)(T_f − 25)

95(95 − T_f) = 2090(T_f − 25)

9025 − 95T_f = 2090T_f − 52250
61275 = 2185 T_f
T_f ≈ 61275 / 2185 ≈ 28.0 °C

So the final equilibrium temperature is about 28.0 °C.
Heat lost by the brass (or gained by the water) ≈ 250·0.380·(95 − 28.0) ≈ 6.37 × 10^3 J.