Specific Heat Capacity

How much energy warms 0.50 kg of water from 20 °C to 100 °C? ($c = 4180\ \text{J kg}^{-1}\text{K}^{-1}$)

1. $E = mc\,\Delta\theta = 0.50 \times 4180 \times 80 = 1.7 \times 10^5\ \text{J}$.

A 0.20 kg brass block ($c = 380\ \text{J kg}^{-1}\text{K}^{-1}$) at 100 °C is dropped into 0.25 kg of water at 20 °C in an insulated cup. Find the final temperature.

1. Energy lost by brass = energy gained by water: $0.20 \times 380 \times (100 - \theta_f) = 0.25 \times 4180 \times (\theta_f - 20)$.
2. $7600 - 76\,\theta_f = 1045\,\theta_f - 20\,900$, so $\theta_f = \dfrac{28\,500}{1121} = 25\ {}^{\circ}\text{C}$.

The water barely warms while the brass falls 75 degrees — that is water's huge $c$ at work.