10 g of ice at $0^{o} C$ absorbs 5460 J of heat energy to melt and change to water at $50^{o} C$ . Calculate the specific latent heat of fusion of ice. Specific heat capacity of water is 4200 J $k g^{-} 1 K^{- 1}$ .

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Solution

Heat gained by ice = 5460 J

$\Rightarrow m L + m s Δ T = 5460$

$\Rightarrow 10 L + 10 \times 4.2 \times 50 = 5460$

$\Rightarrow 10 L = 3360$

$∴ L = 336 J / g$

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Similar questions

10 g of ice at 0 $^{\circ} C$ absorbs 5460 J of heat energy to melt and change t water at 50 $^{\circ} C$ . Calculate the specific latent heat of fusion of ice. (Specific heat capacity of water is 4200 J $K g^{- 1} K^{- 1}$ .

Q. 10 g of ice at

0^{\circ} C

absorbed 5460 J of heat to melt and change into water at

50^{\circ} C

. Calculate the specific latent heat of fusion of ice. Given specific heat capacity of water is 4200

{k g}^{- 1} K^{- 1}

The amount of heat energy required to convert 1 kg of ice at $- 10^{\circ} C$ to water at $100^{\circ} C$ is 7,77,000 J. Calculate the specific latent heat of ice.

Specific heat capacity of ice = 2100 J $k g^{- 1} K^{- 1}$ , specific heat capacity of water=4200 J k $g^{- 1} K^{- 1}$ .

How much heat energy is released when 5.0 g of water at $20^{o} C$ changes into ice at $0^{o} C$ ? Take specific heat capacity of water = 4.2 J $g^{-} 1 K^{- 1}$ , specific latent heat of fusion of ice = 336 J $g^{- 1}$ .