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)`.

A steam at `100^@C` is passed into `1 kg` of water contained in a calo

1.	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» 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`.

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