Steam at `100^(@)C` is passed into `1.1 kg` of water contained in a calorimeter of water equivalent `0.02kg` at `15^(@)C` till the temperature of the calorimeter and its content rises to `80^(@)C`. What is the mass of steam condensed? Latent heat of steam = `536 cal//g`.

Options

A. `.0130`
B. `0.065`
C. `0.260`
D. `0.135`

Answer

A

Explanation

According to principle of calorimetry, heat gained = heat lost. Heat is lost by steam in two stages
(i) Change of state from steam at `100^(@)C` to water at `100^(@)C` is `m xx 540`
(ii) To change water at `100^(@)C` to water at `80^(@)C` is `m xx 1 xx (100-80)`, where m is the mass of the steam condensed. Total heat lost by steam is `m xx 540 + m xx 20= m (540+20)=560m`
Heat gained by calorimeter and its contents is
`(1.1 +0.2) xx (80-15)= 1.12 xx 65` cal
`rArr 560m= 1.12 xx 65`
`rArr m= (1.12 xx 65)/(560)= 0.130kg`