First slide

Calorimetry

Question

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 kg of steam is required per hour?
(Take specific heat of steam = 1 calorie per g°C, Latent heat of vaporization = 540 cal/g)

Moderate

A

1 g
B

1 kg
C

10 g
D

10 kg

Solution

Suppose m kg of steam is required per hour.
Heat is released by steam in following three steps
(i) When $150^{\circ} C steam \underset{Q_{1}}{⟶} 100^{\circ} C steam$
Then $Q_{1} = {mc}_{Steam} Δθ = m \times 1 (150 - 100) = 50 m cal$
(ii) When $100^{\circ} C steam \underset{Q_{2}}{⟶} 100^{\circ} C water$
Then $Q_{2} = {mL}_{V} = m \times 540 = 540 m cal$
(iii) When $100^{\circ} C water \underset{Q_{3}}{⟶} 90^{\circ} C water$
Then $Q_{3} = {mc}_{W} Δθ = m \times 1 \times (100 - 90) = 10 m cal$
Hence, total heat given by the steam,
$Q = Q_{1} + Q_{2} + Q_{3} = 600 m cal$
Hence $Q = Q^{'} \Rightarrow 600 m = 10 \times 10^{3} \times 60$
$\Rightarrow m = 10^{3} g = 1 kg$

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