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Questions based on phase change

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? (Take, specific heat of steam = 1 cal per g⁰C, latent heat of vaporisation = 540 cal g^-1)

Moderate

A

1 g
B

1 kg
C

10 g
D

10 kg

Solution

Suppose m gm steam is required per hour.
Heat is released by steam in following three steps:
(i) When steam at $150^{\circ} C \to steam at 100^{\circ} C$
   $Q_{1} = m c_{steam} Δ θ = m \times 1 (150 - 100) = 50 m cal$
(ii) When steam at    $100^{\circ} C \underset{Q_{2}}{\to} water at 100^{\circ} C$
   $Q_{2} = m L_{V} = m \times 540 = 540 m cal$
(iii) When water at $100^{\circ} C \underset{Q_{2}}{\to} water at 90^{\circ} C$
   $Q_{3} = m c_{w} Δ θ = m \times 1 \times (100 - 90) = 10 m cal$
Hence, total heat given by the steam,
$Q = Q_{1} + Q_{2} + Q_{3} = 600 mcal$ …(i)
Heat taken by 10 kg water,
$Q^{'} = m c_{w} Δ θ = 10 \times 10^{3} \times 1 \times (80 - 20) = 600 \times 10^{3}$ cal
As,    $Q = Q^{'} \Rightarrow 600 m = 600 \times 10^{3} \Rightarrow m = 10^{3} g = 1 kg$

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