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 $=3.35 imes10^5\,\mathrm{J\,kg^{-1}}$ and Latent heat of steam $=2.25 imes10^6\,\mathrm{J\,kg^{-1}}$)

Question

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 $=3.35	imes10^5\,\mathrm{J\,kg^{-1}}$ and Latent heat of steam $=2.25	imes10^6\,\mathrm{J\,kg^{-1}}$)

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $Q=10^{-3}\left[2100(10)+3.35	imes10^5+4180(100)+2.25	imes10^6+1920(10)ight]=10^{-3}(3\,043\,200)\approx3043$ J. Answer (1).

Explanation

Reviewed by Dr. Abubacker

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