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Byju's Answer

Standard IX

Physics

Specific Heat Capacity

In a process,...

Question

In a process, 10 gm of ice at $- 5^{o} C$ is converted into steam at $100^{o} C$ .If latent heat of fusion of ice is 80 cal $g^{- 1}$ , then the amount of heat required to convert 10 g of ice at $0^{\circ} C$ into 10 g of water at same temperature is :

400 c a l

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800 c a l

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1200 c a l

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1600 c a l

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Solution

The correct option is B $800 c a l$
We have ,
$L_{i c e} = 80 c a l / g$
$m = 10 g$
From the definition of latent heat , heat required to convert ice into water at constant temperature $0^{o} C$ ,
$Q = m L_{i c e}$
or $Q = 80 \times 10 = 800 c a l$

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Similar questions

Q. (i) In a process 10 gm of ice at

- 5^{o} C

is converted into steam at

100^{o} C

. If specific heat of water 1 cal

g^{- 1}

^{\circ} C^{- 1}

, the heat required to rise the temperature of water 10 g from

0^{\circ} C

100^{\circ} C

Q. In a process 10 g of ice at

- 5^{o} C

is converted into the steam at

100^{o} C

. If specific heat of ice is

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

, then the amount of heat required to convert 10 g of ice from

- 5^{o} C

0^{o} C

is :

Q. In a process, 10 g of ice at

- 5^{o} C

is converted into the steam at

100^{o} C

. If specific heat of water

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

; the heat required to raise the temperature of 10 g water from

0^{o} C

100^{o} C

is :

Q. The amount of heat required to convert

1 g

of ice at

- 10^{o}

C into steam at

100^{o}

C is (specific heat of ice

= 0.5

cal/g/

^{o}

C, latent heat of ice

= 80

cal/g, latent heat of steam

= 540

cal/g)

Q. The amount of heat required to convert

1

g of ice at

- 10^{0} C

into steam at

100^{0} C

is (approximately): (specific heat of water

=

4.187

J/g

^{0}

C, latent heat of fusion

=

334

J/g , latent heat of vaporization

=

2260

J/g, specific heat of ice

=

2.03

J/g) :

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In a process, 10 gm of ice at −5oC is converted into steam at 100oC.If latent heat of fusion of ice is 80 cal g−1, then the amount of heat required to convert 10 g of ice at 0∘C into 10 g of water at same temperature is :

The correct option is B 800calWe have , Lice=80cal/gm=10gFrom the definition of latent heat , heat required to convert ice into water at constant temperature 0oC , Q=mLiceor Q=80×10=800cal

In a process, 10 gm of ice at $- 5^{o} C$ is converted into steam at $100^{o} C$ .If latent heat of fusion of ice is 80 cal $g^{- 1}$ , then the amount of heat required to convert 10 g of ice at $0^{\circ} C$ into 10 g of water at same temperature is :

The correct option is B $800 c a l$
We have ,
$L_{i c e} = 80 c a l / g$
$m = 10 g$
From the definition of latent heat , heat required to convert ice into water at constant temperature $0^{o} C$ ,
$Q = m L_{i c e}$
or $Q = 80 \times 10 = 800 c a l$