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The heat energy required to convert 10 g of ice at $0^{0} C$ to water at $20^{0} C$ is (Latent heat of fusion of ice $= 80 {cal g}^{- 1},$ specific heat capacity of water $= 1 {cal g}^{- 1}^{0} C^{- 1},$ mechanical equivalent of heat $= 4.2 {J cal}^{- 1})$

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$4200 J$

$2100 J$

$8400 J$

$1000 J$

answer is A.

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Detailed Solution

Heat required is

$H = M \times L + M S ∆ T$

$H = 10 \times 80 \times 4.2 + 10 \times 1 \times 4.2 \times 20 = 4200 J$

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