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 $= 2100\ \text{J kg}^{-1}\text{K}^{-1}$, specific heat of water $= 4180\ \text{J kg}^{-1}\text{K}^{-1}$, specific heat of steam $= 1920\ \text{J kg}^{-1}\text{K}^{-1}$, latent heat of ice $= 3.35\times10^{5}\ \text{J kg}^{-1}$ and latent heat of steam $= 2.25\times10^{6}\ \text{J kg}^{-1}$)
Answer: (A) $3043\ \text{J}$
With $m = 10^{-3}\ \text{kg}$, add the heat for each stage:
- ice $-10^\circ\text{C} \to 0^\circ\text{C}$: $10^{-3}\times2100\times10 = 21\ \text{J}$
- melting: $10^{-3}\times3.35\times10^5 = 335\ \text{J}$
- water $0 \to 100^\circ\text{C}$: $10^{-3}\times4180\times100 = 418\ \text{J}$
- boiling: $10^{-3}\times2.25\times10^6 = 2250\ \text{J}$
- steam $100 \to 110^\circ\text{C}$: $10^{-3}\times1920\times10 = 19.2\ \text{J}$
Total $= 21 + 335 + 418 + 2250 + 19.2 = 3043.2 \approx 3043\ \text{J}$
Solution by Sreeraj P, M.Sc Physics