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Standard XII

Physics

Latent Heat

150 g of ice ...

Question

150 g of ice is mixed with 100 g of water at temperature $80^{o} C$ . The latent heat of ice is 80 cal/g and the specific heat of water is 1 cal/g $-^{o} C$ . Assuming no heat loss to the environment, the amount of ice which does not melt is

100 g

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0 g

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150 g

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50 g

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Solution

The correct option is B 50 g
Heat produced by water is $Q = 100 \times 1 \times 80 = 8000 c a l$
Ice melts= $\frac{8000}{80} = 100 g m$
ice remaing=150-100=50gm

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Similar questions

150 g

of ice is mixed with

100 g

of water at temperature

80^{o} C

. The latent heal of ice is

80 c a l / g

and the specific heat of water is

1 c a l / g^{o} C

. Assuming no heat loss to the environment, the amount of ice which does not melt is :

Q. The amount of heat required to raise the temperature of

75 g

of ice at

0^{o} C

to water at

10^{o} C

is:

[latent heat of fusion of ice is

80 c a l / g

, specific heat of water is

1 c a l / g^{o} C

]

Q. 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.

2

kg ice at

- 20^{o} C

is mixed with

5

kg water at

20^{o} C

. The final amount of water in the mixture would be ____.
Given the specific heat of ice =

0.5 c a l / g^{C}

; Specific heat of water =

1 c a l / g^{C}

; Latent heat of fusion for ice =

80 c a l / g

10 g

of ice at

0^{o} C

is mixed with

100 g

of water at

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in a calorimeter. The final temperature of the mixture is [Specific heat of water

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, latent heat of fusion of ice

= 80 c a l g^{- 1}]

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150 g of ice is mixed with 100 g of water at temperature 80oC. The latent heat of ice is 80 cal/g and the specific heat of water is 1 cal/g −oC . Assuming no heat loss to the environment, the amount of ice which does not melt is

The correct option is B 50 gHeat produced by water is Q=100×1×80=8000calIce melts=800080=100gmice remaing=150-100=50gm

150 g of ice is mixed with 100 g of water at temperature $80^{o} C$ . The latent heat of ice is 80 cal/g and the specific heat of water is 1 cal/g $-^{o} C$ . Assuming no heat loss to the environment, the amount of ice which does not melt is

The correct option is B 50 g
Heat produced by water is $Q = 100 \times 1 \times 80 = 8000 c a l$
Ice melts= $\frac{8000}{80} = 100 g m$
ice remaing=150-100=50gm