Question

A steam at `100^@C` is passed into `1 kg` of water contained in a calorimeter of water equivalent `0.2 kg` at `9^@C` till the temperature of the calorimeter and water in it is increased to `90^@C`. Find the mass of steam condensed in `kg(S_(w) = 1 cal//g^@C, & L_("steam") = 540 cal//g) (EAM = 14 E)`.

Answer 1

Let, `m` be the mass of the steam condensed.
Mass of the steam passed into calorimeter,
`m_(2) = 1 kg = 1000 g`.
Water equivalent of calorimeter,
`m_(1)S_(1) = 0.2 kg = 200 kg`
`theta_(1)` = temperature of the steam `= 100^@ C`
`theta_(2)` = temperature of the water `= 9^@C`
`theta_(3)` = resultant temperature `= 90^@C`
Heat lost = heat gained (calorimeter + water)
`m[L_(steam) + S_(W) (theta_(1) -theta_(3))] = [m_(1)S_(1) + m_(2) S_(W)] (theta_(3) -theta_(2))`
`m[540+1(100-90)] = [200 + 1000 xx 1](90-9)`
`rArr m = 176 = 0.176 kg~~ 0.18 kg`.