Question

Question #1a163

Answer 1

`154500cal`

Explanation:

We know that 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.

In the given problem heat is given to water to increase its temperature from `22^@"C"` to `100^@"C"` and thereafter boil the water to make it steam at `100^@"C"`.

1. Water heated from `22^@"C"` to `100^@"C"`
   Using the expression and taking specific heat of water as 1.
   `DeltaQ=mst`
   `DeltaQ_1=250xx1xx(100-22)=19500cal`

2. Heat gained by boiling water to change into vapours at the same temperature i.e., `100^@"C"` is given by `DeltaQ_2=mL`, where Latent heat of vaporization of water is `540calg^-1`.
   `:. DeltaQ_2=250xx540=135000cal`

Total heat required `=DeltaQ_1+DeltaQ_2=19500+135000`
`=154500cal`