1. Heat required to convert ice at $-10^\circ\text{C}$ to steam at $110^\circ\text{C}$

We add heat in five stages:

• heat ice from $-10^\circ\text{C}$ to $0^\circ\text{C}$
• melt ice at $0^\circ\text{C}$
• heat water from $0^\circ\text{C}$ to $100^\circ\text{C}$
• convert water at $100^\circ\text{C}$ to steam
• heat steam from $100^\circ\text{C}$ to $110^\circ\text{C}$

Given:

$m=10^{-3}\,\text{kg}$ $c_{\text{ice}}=2100\,\text{J kg}^{-1}\text{K}^{-1}$ $c_{\text{water}}=4180\,\text{J kg}^{-1}\text{K}^{-1}$ $c_{\text{steam}}=1920\,\text{J kg}^{-1}\text{K}^{-1}$ $L_f=3.35\times 10^5\,\text{J kg}^{-1}$ $L_v=2.25\times 10^6\,\text{J kg}^{-1}$

2. Stage 1: Heating ice from $-10^\circ\text{C}$ to $0^\circ\text{C}$

$Q_1=mc_{\text{ice}}\Delta T$ $Q_1=(10^{-3})(2100)(10)=21\,\text{J}$

3. Stage 2: Melting the ice at $0^\circ\text{C}$

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

4. Stage 3: Heating water from $0^\circ\text{C}$ to $100^\circ\text{C}$

$Q_3=mc_{\text{water}}\Delta T$ $Q_3=(10^{-3})(4180)(100)=418\,\text{J}$

5. Stage 4: Converting water at $100^\circ\text{C}$ to steam

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

6. Stage 5: Heating steam from $100^\circ\text{C}$ to $110^\circ\text{C}$

$Q_5=mc_{\text{steam}}\Delta T$ $Q_5=(10^{-3})(1920)(10)=19.2\,\text{J}$

7. Total heat required

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

So, approximately,

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

8. Matching with options

• A: $3022\,\text{J}$
• B: $3043\,\text{J}$
• C: $3003\,\text{J}$
• D: $3024\,\text{J}$

Therefore, the correct option is B.

Note: The question says "work required," but in this thermodynamics context it clearly means the heat energy required for the full conversion.