Question

In an industrial process 10 kg of water per hour is to be heated from 20⁰C to 80⁰C. To do this steam at 150⁰C 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 90⁰C. how many kg of steam is required per hour.

(Specific heat of steam = 1 calorie per gm⁰C, Latent heat of vaporisation = 540 cal/gm)

• A. 1 gm
• B. 1 kg
• C. 10 gm
• D. 10 kg

Answer 1

Answer: (B)

1 kg

Answer 2

Heat required by 10 kg water to change its temperature from 20⁰C to 80⁰C in one hour is

Q₁ = $(mc\Delta T)_{water}$ =

$(10 \times 10^{3}) \times 1 \times (80–20) = 600 \times 10^{3}calorie$In condensation (i) Steam release heat when it looses it's temperature from 150⁰C to 100⁰C. $\lbrack mc_{steam}\Delta T\rbrack$

(ii) At 100⁰C it converts into water and gives the latent heat. $\lbrack mL\rbrack$

(iii) Water release heat when it looses it's temperature from 100⁰C to 90⁰C. $\lbrack ms_{water}\Delta T\rbrack$

If m gm steam condensed per hour, then heat released by steam in converting water of 90⁰C

Q₂=$(mc\Delta T)_{steam} + mL_{steam} + (ms\Delta T)_{water}$

= $m\lbrack 1 \times (150 - 100) + 540 + 1 \times (100 - 90)\rbrack$ = 600 m calorie

According to problem Q₁ = Q₂ ⇒ 600 × 10³ cal = 600 m cal ⇒ m = 10³ gm = 1 kg.