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