Question

Thinking Question on Latent Heat, Help?

Answer 1

Final temperature is `100^@"C"`
Mixture is `1263.bar8gm` of water and `736.bar1gm` of steam at the above temperature.

Explanation:

We know that in such interactions heat is gained by one and heat is lost is by the other.

Also that the heat gained/lost is given by
`DeltaQ=mst`, or `DeltaQ=mL`
where `m,s and t` are the mass, specific heat and rise or gain in temperature of the object;
`L` is the latent heat for the change of state and
`Delta Q_"lost"=Delta Q_"gained"`

In the given problem heat is lost by steam and gained by ice.
We know that latent heat of fusion of water `=80 calgm^-1` is much smaller than latent heat of steam `=540 calgm^-1`. Specific heat capacity of water is taken as `1`, and of ice and steam `0.5`

In the given question ice is at `-25^@"C"` and steam at `200^@"C"`. Mass of ice and steam being same. It is difficult to assess what is going to be the final mixture. therefore, we proceed step by step.

Step. 1. It is clear that ice will melt and will become water at `0^@"C"`.

`"Heat gained by ice at " -25^@"C to change in to ice at " 0^@"C"`
`=1000xx0.5xx25=12500cal` .....(1)

On the other hand we have `"Heat lost by steam at " 200^@"C to change in to steam at " 100^@"C"`
`=1000xx0.5xx100=50000cal` ....(2)

Comparing (1) and (2) we assess that we have excess heat from (2) which can be given to ice to melt it.

`:. "Heat gained by ice at "0^@"C to change into water at "0^@"C"`
`=1000xx80=80000cal` .....(3)

Now comparing heat lost with heat gained in (1), (2) and (3), the required heat for melting of ice will be provided by change of state of steam.
Difference between the heat lost and gained `=80000+12500-50000=42500cal`

Step 2. `"Heat gained by water at "0^@"C to change in to water at "` `100^@"C"=1000xx1xx100=100000` ......(4)

Now the difference between heat gained and heat lost from (1) to (4) `=80000+12500+100000-50000=142500cal`
This can be supplied by `142500/540=263.bar8gm` of steam by its change of state.

The remaining steam at `100^@"C"=1000-263.bar8=736.bar1gm`

Since both the components are now at the same temperature, there is no further exchange of heat.