**31.** An amount of ice of mass \( 10^{-3} \, \text{kg} \) and temperature \( -10^\circ \text{C} \) is transformed to vapour of temperature \( 110^\circ \text{C} \) by applying heat. The total amount of work required for this conversion is: (Take: Specific heat of ice = \( 2100 \, \text{Jkg}^{-1}\text{K}^{-1} \), Specific heat of water = \( 4180 \, \text{Jkg}^{-1}\text{K}^{-1} \), Specific heat of steam = \( 1920 \, \text{Jkg}^{-1}\text{K}^{-1} \), Latent heat of ice = \( 3.35 \times 10^5 \, \text{Jkg}^{-1} \), Latent heat of steam = \( 2.25 \times 10^6 \, \text{Jkg}^{-1} \)) Options: - (1) 3022 J - (2) 3043 J - (3) 3003 J - (4) 3024 J

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🧊➡️💧➡️💨 Turning ice-cold ice into hot steam

Imagine you have a tiny ice cube that is really cold ( −10 °C ). You want to turn it into hot steam ( 110 °C ).
To do that you must keep adding heat while the ice passes through several stages:

1. Warm the ice so it is not so cold.
2. Melt the ice into water — this needs extra energy called latent heat.
3. Warm the water up to boiling point.
4. Boil the water into steam — again a latent heat step.
5. Warm the steam a little more. Add the energy for every step and you get the total heat you must supply.
   Because the ice piece is very tiny (only 1 g), the final answer is only about 3 kilo-Joules of energy.

Given Data

Mass: $m = 10^{-3}\,\text{kg}$

Specific heats
Ice: $c_i = 2100\,Jkg^{-1}K^{-1}$
Water: $c_w = 4180\,Jkg^{-1}K^{-1}$
Steam: $c_s = 1920\,Jkg^{-1}K^{-1}$

Latent heats
Fusion (ice): $L_f = 3.35\times10^{5}\,Jkg^{-1}$
Vaporisation (steam): $L_v = 2.25\times10^{6}\,Jkg^{-1}$

Temperature changes
Ice: $-10^{\circ}C \to 0^{\circ}C \Rightarrow \Delta T_1 = 10\,K$
Water: $0^{\circ}C \to 100^{\circ}C \Rightarrow \Delta T_3 = 100\,K$
Steam: $100^{\circ}C \to 110^{\circ}C \Rightarrow \Delta T_5 = 10\,K$

Step-wise Heat Calculations

1. Warm the ice $Q_1 = m c_i \Delta T_1 = (10^{-3})\,(2100)\,(10) = 21\,J$

2. Melt the ice $Q_2 = m L_f = (10^{-3})\,(3.35\times10^{5}) = 335\,J$

3. Warm the water $Q_3 = m c_w \Delta T_3 = (10^{-3})\,(4180)\,(100) = 418\,J$

4. Vaporise the water $Q_4 = m L_v = (10^{-3})\,(2.25\times10^{6}) = 2250\,J$

5. Warm the steam $Q_5 = m c_s \Delta T_5 = (10^{-3})\,(1920)\,(10) = 19.2\,J$

Total Heat / Work Required