Question

100 g of ice is contained in a thermos flask with a minimal heat capacity. Determine how much steam, at a temperature of 100°C, is required to simply melt the ice. Given that the specific heat capacity of water is 4.2 J g⁻¹ °C⁻¹ and the latent heat of ice is 336 J g⁻¹.

Answer 1

Heat gained by ice to melt at 0°C:

Q_gain = M_ice × L_ice

= 100 × 336

= 33600 J

The steam at 100°C first condenses into water at 100°C, and then this water cools down to the final mixture temperature of 0°C (since the ice is simply melting). Let the mass of steam be m_s:

Q_loss = (m_s × L_steam) + (m_s × c_w × ΔT)

= (m_s × 2260) + (m_s × 4.2 × [100 − 0])

= 2260 m_s + 420 m_s

= 2680 m_s

By Principle of Calorimetry:

Q_gain = Q_loss

33600 = 2680 m_s

m_s = `33600/2680`

m_s = 12.54 g