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Standard XII

Physics

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 $80 c a / g$ 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

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 A $50 g$
Heat given by water $= 100 \times 1 \times 800 = 8000 c a l$
Heat taken by ice $= 8000 c a l = m \times 80$
On solving we get $m = 100 g m$
So amount of ice which does not melt $= 150 - 100 = 50 g m$ .

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

The correct option is A 50gHeat given by water =100×1×800=8000calHeat taken by ice =8000cal=m×80On solving we get m=100gmSo amount of ice which does not melt =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 $80 c a / g$ 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

The correct option is A $50 g$
Heat given by water $= 100 \times 1 \times 800 = 8000 c a l$
Heat taken by ice $= 8000 c a l = m \times 80$
On solving we get $m = 100 g m$
So amount of ice which does not melt $= 150 - 100 = 50 g m$ .