Work done in converting one gram of ice at −10∘C into steam at 100∘C is -[Given, specific heat of ice ci=0.5 cal g−1 ∘C−1, specific heat of water cw=1 cal g−1 ∘C−1, latent heat of fusion of ice Lf=80 cal/g, & latent heat of vaporization of steam Lv=540 cal/g]

Work done in converting one gram of ice at −10∘C into steam at 100

1.	Work done in converting one gram of ice at −10∘C into steam at 100∘C is -[Given, specific heat of ice ci=0.5 cal g−1 ∘C−1, specific heat of water cw=1 cal g−1 ∘C−1, latent heat of fusion of ice Lf=80 cal/g, & latent heat of vaporization of steam Lv=540 cal/g]
Answer» Work done in converting one gram of ice at $- 10^{\circ} C$ into steam at $100^{\circ} C$ is - [Given, specific heat of ice $c_{i} = 0.5 {cal g}^{- 1}^{\circ} C^{- 1}$ , specific heat of water $c_{w} = 1 {cal g}^{- 1}^{\circ} C^{- 1}$ , latent heat of fusion of ice $L_{f} = 80 cal/g$ , & latent heat of vaporization of steam $L_{v} = 540 cal/g$ ]

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Work done in converting one gram of ice at −10∘C into steam at 100∘C is -[Given, specific heat of ice ci=0.5 cal g−1 ∘C−1, specific heat of water cw=1 cal g−1 ∘C−1, latent heat of fusion of ice Lf=80 cal/g, &amp; latent heat of vaporization of steam Lv=540 cal/g]

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Work done in converting one gram of ice at −10∘C into steam at 100∘C is -[Given, specific heat of ice ci=0.5 cal g−1 ∘C−1, specific heat of water cw=1 cal g−1 ∘C−1, latent heat of fusion of ice Lf=80 cal/g, & latent heat of vaporization of steam Lv=540 cal/g]