If 100 g of ice at 0C is mixed with 100 g of boiling water at 100C

Graph the variation of temperature(T) vs time (t) of the two components of the mixture?

### 2 Answers | Add Yours

When 100 g of ice at `0^oC` is mixed with 100 g of boiling water at `100^o C` , heat is transferred from the boiling water to the ice, resulting its melting into water at `0^oC ` and then its temperature will increase gradually till it reaches a thermal equilibrium.

Let the final temperature of the mixture, after thermal equilibrium is established, be `x^o` C.

Amount of heat absorbed in melting the ice=latent heat* mass

`=80*100`

`=8000` cal (latent heat of fusion of ice = 80 cal/g)

Amount of heat absorbed in raising the temperature of 100 g water at 0^C to x^oC=mass*specific heat*temperature rise

`=100*1*x` (specific heat of water is 1 cal/g)

=`100x` cal

Total heat absorbed=`(8000+100x)` cal

This heat will be supplied by the boiling water.

Heat released by its cooling to `x^oC`

`=100*1*(100-x)`

=10000-100x

By the condition of the equilibrium,

`(8000+100x)=(10000-100x)`

`rArr 200x=2000`

`rArr x=10`

So, final temperature of the mixture is `10^oC`

The graph is shown in the attached figure.

*Addendum:*

The other component of the mixture will exhibit a variation as depicted in the attached image.

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes