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 vaporization`=540 cal//g`)?

Options

A. 1 g
B. 1 kg
C. 10 g
D. 10 kg

Answer

B

Explanation

Suppose m kg steam is required per hour
Heat is released by steam in following three steps:
i) When steam at `150^(@)`C tosteam at `100^(@)`C
`Q_(1)=mc_(steam)Deltatheta)`
`m xx1(150-100) = 50 mcal`
ii) When steam at `100^9@)`C `to` water at `100^(@)`C
`Q_(2)=mL_(V)=mxx540`=540m cal
iii) When water at `100^(@)`C to `water at `90^(@)`C
`Q_(3) = mc_(W)Deltatheta`
`=m xx 1 xx (100-90) = 10m` cal
Heat, total heat given by the steam
Q = `Q_(1)+Q_(2)+Q_(3)`= 600m cal
Heat taken by 10 kg water
`Q^(') = mc_(W)deltatheta`
`=10 xx 10^(3)xx1xx(80-20)`
`=600 xx 10^(3)` cal
Hence, `Q=Q^(')`
`rArr` 600m = `600 xx 10^(3)`
`rArr m = 10^(3)`g = 1kg