In an industrial process 10 kg of water per hour is to be heated from `20^@C` to `80^@C`. To do this steam at `150^@C` is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at `90^@C`. How many kilograms of steam is required per hour (specific heat of steam`=1 cal//g^@C`, Latent heat of vapourization`=540 cal//g`)?

Answer

1

Explanation

Suppose m kg steam is required per hour lthrgt Heat is released by steam in following three steps
(i) When `150^(@)C` steam `underset(Q_(1))rarr 100^(@)C` steam `" " Q_(1) = "mc"_("steam") Deltatheta = "m" xx1 (150-100) = 50"m cal"`
(ii) When `100 ^(@)C` steam `underset(Q_(2))rarr 100^(@)C` water `" " Q_(2) ="m"L_("v") = "m" xx540 = 540 "m" "cal"`
(iii) When `100 ^(@)C` water `underset(Q_(2))rarr 90^(@)C`water `" " Q_(3) =mc_("w")Deltatheta = m xx 1 xx (100-90) = 10 m"cal"`
Hence total heat given by the steam `" " Q = Q_(1) + Q_(2) + Q_(3) = 600 m "cal" ........(i)`
Heat taken by `10` kg water `Q' mc_(w)Deltatheta = 10 xx 10^(3) xx 1 xx (80-20) = 600 xx 10^(3)"cal"`
Hence `Q=Q' implies 600 "m" = 600 xx 10^(3) "m" = 10^(3) "gm" = 1 kg`