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Question

Latent heat of fusion of ice is:
$336 k J k g^{- 1}$
$663 k J k g^{- 1}$
$363 k J k g^{- 1}$
$636 k J k g^{- 1}$

A

336 k J k g^{- 1}

B

363 k J k g^{- 1}

C

636 k J k g^{- 1}

D

663 k J k g^{- 1}

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Solution

Verified by Toppr

Latent heat of fusion of ice
$= 80 c a l g^{- 1}$
$= 80 \times 4.2 J \times 10^{3} k g^{- 1}$
$= 336 \times 10^{3} J k g^{- 1}$

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Similar Questions

Q1

Assertion :Latent heat of fusion of ice is

336000 J k g^{- 1}

. Reason: Latent heat refers to change of state without any change in temperature.

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Q2

400 g of ice at 253 K is mixed with 0.05 kg of steam at

100^{o} C

. Latent heat of vaporisation of steam

= 540 c a l g^{- 1}

. Latent heat of fusion of ice

= 80 c a l g^{- 1}

. Specific heat of ice

= 0.5 c a l g^{- 1}^{o} C^{- 1}

. Find the resultant temperature of the mixture.

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Q3

Work done in converting one gram of ice at

- 10^{\circ} C

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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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Q4

Calculate the amount of heat reqiured to convert

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Q5

What is the entropy change in

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