Steam at 100°C is added to ice at 0°C. Find the temperature when the mass of steam is 10 g and
the mass of ice is 50 g. The specific heat of water is 4186 J/kg°C, its latent heat of fusion is 3.33x105 J/kg and its heat of vaporization is 2.26x106 J/kg.

The final temperature of the mixture of steam and ice is 40.36 ⁰C.

The given parameters:

The final temperature of the mixture of steam and ice is calculated as follows;

Thus, the final temperature of the mixture of steam and ice is 40.36 ⁰C.

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Explanation:

First of all, we have to calculate the heat released by the steam whem it turns into liquid water. This is given by:

where

is the mass of steam

is the latent heat of vaporization

Substituting,

Now we find the heat absorbed by the ice when it turns into liquid water:

where

is the mass of ice

is the latent heat of fusion

So,

Then, we call:

the heat absorbed by the ice (now liquid water) till reaching the equilibrium temperature T, where

is the specific heat capacity of water

the heat released by the steam (now liquid water) till reaching the equilibrium temperature T

Since the total heat absorbed must be equal to the total heat released, we have:

And now we can solve to find T: