An amount of ice of mass $$10^{-3}kg \text{ and temperature } -10^{o}C$$ is transformed to vapour of temperature $$110^{o}C$$ by applying heat. The total amount of work required for this conversion is, (Take, specific heat of ice $$= 2100Jkg^{-1}K^{-1},$$ specific heat of water $$ 4180Jkg^{-1}K^{-1},$$  specific heat of steam  $$=1920Jkg^{-1}K^{-1},$$  Latent heat of ice $$=3.35\times10^{5}Jkg^{-1} $$ and Latent heat of steam $$ = 2.25\times10^{6}Jkg^{-1})$$

We need to find the total heat required to convert ice at $$-10°C$$ to steam at $$110°C$$.

Given data:

Mass $$m = 10^{-3}$$ kg, specific heat of ice $$c_{ice} = 2100 \text{ J kg}^{-1}\text{K}^{-1}$$, specific heat of water $$c_w = 4180 \text{ J kg}^{-1}\text{K}^{-1}$$, specific heat of steam $$c_s = 1920 \text{ J kg}^{-1}\text{K}^{-1}$$, latent heat of ice $$L_f = 3.35 \times 10^5 \text{ J kg}^{-1}$$, latent heat of steam $$L_v = 2.25 \times 10^6 \text{ J kg}^{-1}$$.

$$Q_1 = m \cdot c_{ice} \cdot \Delta T = 10^{-3} \times 2100 \times 10 = 21 \text{ J}$$

$$Q_2 = m \cdot L_f = 10^{-3} \times 3.35 \times 10^5 = 335 \text{ J}$$

$$Q_3 = m \cdot c_w \cdot \Delta T = 10^{-3} \times 4180 \times 100 = 418 \text{ J}$$

$$Q_4 = m \cdot L_v = 10^{-3} \times 2.25 \times 10^6 = 2250 \text{ J}$$

$$Q_5 = m \cdot c_s \cdot \Delta T = 10^{-3} \times 1920 \times 10 = 19.2 \text{ J}$$

$$Q = Q_1 + Q_2 + Q_3 + Q_4 + Q_5 = 21 + 335 + 418 + 2250 + 19.2 = 3043.2 \text{ J}$$

$$Q \approx 3043 \text{ J}$$

The correct answer is Option 1: 3043 J.