10 g of ice at 0-C absorbs 5460 J of heat energy to melt and change to water at 50-C- Calculate the specific latent heat of fusion of ice- Specific heat capacity of water is 4200 J kg-1 K-1-

The correct option is B 336 Jg^ 1 Heat absorb = mL+mc× (T 0) 5460=10× L+10× 4.2× 50 L=336Jg^ 1 ...

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Byju's Answer

Standard IX

Chemistry

Latent Heat of Vaporisation

10 g of ice a...

Question

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

2168 J g^{- 1}

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336 J g^{- 1}

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43.36 J g^{- 1}

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21.68 J g^{- 1}

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Solution

The correct option is B $336 J g^{- 1}$
Heat absorb = $m L + m c \times (T - 0)$
$5460 = 10 \times L + 10 \times 4.2 \times 50$
$L = 336 J g^{- 1}$

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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}

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 $g^{- 1}$ .

Q. The amount of heat energy required to convert

1

kg of ice at

- 10^{\circ} C

to water at

100^{\circ} C

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}

A refrigerator converts 100 g of water at $20^{o} C$ to ice at $- 10^{o} C$ in 73.5 min. Calculate the average rate of heat extraction in watt. The specific heat capacity of water is 4.2 J $g^{- 1} K^{- 1}$ , specific latent heat of ice is 336 J $g^{-} 1$ and the specific heat capacity of ice is 2.1 J $g^{- 1} K^{- 1}$ .

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10 g of ice at 0∘C absorbs 5460 J of heat energy to melt and change to water at 50∘C. Calculate the specific latent heat of fusion of ice. Specific heat capacity of water is 4200Jkg−1K−1.

The correct option is B 336Jg−1Heat absorb =mL+mc×(T−0)5460=10×L+10×4.2×50L=336Jg−1

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

The correct option is B $336 J g^{- 1}$
Heat absorb = $m L + m c \times (T - 0)$
$5460 = 10 \times L + 10 \times 4.2 \times 50$
$L = 336 J g^{- 1}$