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

When `150^@` C steam ` overset(Q_1)rarr 100^(@)C` steam
`Q_1 = mS_(s) Delta theta = m xx 1 xx (150 - 100) = 50 m cal`
When `100^@ C` steam `overset(Q_2) rarr 100^(@)C` water
`Q_2 = mL_(v) = m xx 540 = 540 m Cal`
`100^(@)C` water `overset(Q_3) rarr 90^@ C` water
`Q_(3) = mS_(w) Delta theta = mxx 1 (100-90) = 10 m cal`
Now,
`Heat loss = heat gain
`rArr 50 m + 540 m + 10 m =m S_w Delta theta`
`rArr 600 xx m = 10 xx 10^(3) xx 1 (80 - 20)`
`rArr m = 10^(3) gm = 1 kg.`