The temperature of 100g of water is to be raised from `24^@C` to `90^@C` by adding steam to it. Calculate the mass of the steam required for this purpose.

Answer

12

Explanation

Let m be the mass of the steam required to raise the temperature of 100 g of water from `24^(@)C` to `90^(@)C` . Heat lost by steam = Heat gained by water
`thereforem(L+sDeltatheta_(1))=100sDeltatheta_(2) or m=((100)(s)(Deltatheta_(2)))/(L+s(Deltatheta_(1)))`
Here, s = specific heat of water = 1 cal/g-°C,
L = latent heat of vaporisation = 540 cal/g.
`Deltatheta_(1)=(100-90)=10^(@)CandDeltatheta_(2)=(90-24)=66^(@)C`
Substituting the values, we have `m=((100)(1)(66))/((540)+(1)(10))=12grArrthereforem=12g`