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`)?

Options

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

Answer

B

Explanation

Heat required by 10 kg water to change its temperature from `20^@C` to `80^@C` in one hour is
`Q_1=(mcDeltaT)_("water")=(10xx10^3)xx1xx(80-20)=600xx10^3` cal
In condensation
i. Steam releases heat when it loses its temperature from `150^@C` to `100^@C`.
ii. At `100^@C` it converts into water and gives the latent heat.
iii. Water releases heat when it loses its temperature from `100^@C` to `90^@C`.
If m grams of steam is condensed per hour, then heat released by steam in converting to water at `90^@C`
`Q_2=(mcDeltaT)_("steam")+mL_("steam")+(msDeltaT)_("water")`
`=m[1xx(150-100)+540+1xx(100-90)]=600mcal`
According to problem, `Q_1=Q_2implies600xx10^3=600` m cal
`implies m=10^3g=1kg`