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

To calculate the total amount of heat required to transform $10^{-3} \, \text{kg}$ of ice at $-10^\circ\text{C}$ into steam at $110^\circ\text{C}$, we break the process into steps:

Step 1: Heat required to raise the temperature of ice from $-10^\circ\text{C}$ to $0^\circ\text{C}$

$$Q_1 = m \cdot c_{\text{ice}} \cdot \Delta T$$

 $$Q_1 = (10^{-3}) \cdot (2100) \cdot (10) = 21 \, \text{J}$$ 

Step 2: Heat required to melt the ice at $0^\circ\text{C}$

$$Q_2 = m \cdot L_{\text{fusion}}$$

 $$Q_2 = (10^{-3}) \cdot (3.35 \times 10^5) = 335 \, \text{J}$$ 

Step 3: Heat required to raise the temperature of water from $0^\circ\text{C}$ to $100^\circ\text{C}$

$$Q_3 = m \cdot c_{\text{water}} \cdot \Delta T$$

 $$Q_3 = (10^{-3}) \cdot (4180) \cdot (100) = 418 \, \text{J}$$ 

Step 4: Heat required to convert water to steam at $100^\circ\text{C}$

$$Q_4 = m \cdot L_{\text{vaporization}}$$

 $$Q_4 = (10^{-3}) \cdot (2.25 \times 10^6) = 2250 \, \text{J}$$ 

Step 5: Heat required to raise the temperature of steam from $100^\circ\text{C}$ to $110^\circ\text{C}$

$$Q_5 = m \cdot c_{\text{steam}} \cdot \Delta T$$

 $$Q_5 = (10^{-3}) \cdot (1920) \cdot (10) = 19.2 \, \text{J}$$ 

Total Heat Required:

$$Q_{\text{total}} = Q_1 + Q_2 + Q_3 + Q_4 + Q_5$$

 $$Q_{\text{total}} = 21 + 335 + 418 + 2250 + 19.2 = 3043.2 \, \text{J}$$