Question

1 gram of ice is passed into container having 2 grams of water at 35 degree Celsius the equilibrium temperature of mixture is

Answer 1

Explanation:

Total heat gained by ice is equal to the total heat lost by steam.

For ice to completely convert into water, heat required is m

1

L

f

=1×80=80cal

For steam to completely convert into water, heat released is m

2

L

v

=1×540=540 cal

Hence, first 80 calories will not be enough for the steam to condense completely.

Now, to convert melted water to 100

o

C from 0

o

C, heat required is m

1

s(100−0)=1×1×100=100cal

So, total energy required to heat ice to water 100

o

C is 100+80=180 cal.

Hence, even this amount of energy is not enough for the steam to condense completely. Hence, the final temperature of the mixture will be 100

o

C.

Note- finally the mixture will consist of both steam and water at 100

o

C.

Answer 2

The equilibrium temperature will be 0°C

Total heat available when 2 grams of water cools down from 35°C will be

2× 1 ×( 35-0) = 70 calories

The heat that 1 gram of ice needs to completely melt will be

1×80 = 80 calories

But heat available will be only that which is liberated from water

that means all the ice will not melt and some ice will be present in the end.

Therefore, the equilibrium temperature will be 0°C