Question

Calculate the number of joules given off when 32.0 grams of steam cools from 110.0 °C to ice at -40.0 °C.

Answer 1

First the steam should cool to 100°C, next it should turn into water and cool to 0°C, then the water should turn into ice and the ice should cool to -40.0°C. Let us write the amount of energy corresponding to each stage of process.

1) $Q_1 = c_s m_s \Delta T_s = 2078\cdot 0.032\cdot 10°\mathrm{C} = 665\,\mathrm{J}$

2) $Q_2 = r m_s = 2.26\cdot10^6\cdot0.032= 72320\,\mathrm{J}$

3) $Q_3 = c_w m_w \Delta T_w = 4200\cdot0.032\cdot100°\mathrm{C} = 13440\,\mathrm{J}$

4) $Q_4 = \lambda m_w = 330000\cdot0.032 = 10560\,\mathrm{J}$

5) $Q_5 = c_im_i\Delta T_i = 2110\cdot0.032\cdot40°\mathrm{C} = 2701\,\mathrm{J}.$

The total amount of energy will be $665+72320+13440+10560+2701 =99686\,\mathrm{J}\approx 10^5\,\mathrm{J}.$