When 1g of ice ai 0'C is mixed with 1g of steam at 100'C then the temperature of the mixture at the equillibrium is...

I am going to assume that both the steam and ice at at their respective temperatures to change state, ie 100C and 0C.

I am also going to assume that you have knowledge of

1. latent heat of fusion and vaporization
2. specific heat capacity
3. Specific heat capacity of ice, water and steam are different.

And we need additional information as well. Luckily for us, we can ask Google. For water the latent heat of heat is 334J/g and the latent heat of vaporization is 2230J/g. Specific heat capacity of water is 4.2J/(gK).

The steam condenses into 100C water, releasing 2230J. 334J of it goes to melting the ice into 0C water. The rest (1896J) heats up the 0C water to 1896/4.2 = 451C. But water cannot be heated up to 451C! What this means is that not all the steam needs to condense to bring the 0C ice to 100C water. And therefore the water/steam mixture is at 100C.

Regards

Arun (askIITians forum expert)