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Standard XII

Physics

Emissive and Absorptive Power

Steam at 100∘...

Question

Steam at $100^{o} C$ is passed through 1.5 kg water which has $20^{o} C$ temperature kept in a container. Water equivalent of container is 0.50 kg and final temperature of mixture is $80^{o} C$ . Find the final mass (approximate) of water in container.
$(L_{v a p} = 540 c a l / g, s_{w a t e r} = 1 \frac{c a l}{g -^{o} C})$

1.7 kg

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Solution

The correct option is A 1.7 kg
Heat will be added to both water and container, since water equivalent of container is given, therefore it can directly be added in mass of water for heat added. Therefore,
$m = 1.5 + 0.5 = 2 k g$
Now,
$m s △ θ = m L + m s △ θ$
$2 \times 1 \times 60 = m \times 540 + m \times 1 \times 20$
$120 = m (560)$
$m_{s t e a m} = \frac{120}{560} = \frac{3}{14} = .214 k g$
$m_{t o t a l} = m_{w a t e r} + m_{s t e a m} = 1.5 + 0.21 = 1.71 k g ≃ 1.7 k g$

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Steam at 100oC is passed through 1.5 kg water which has 20oC temperature kept in a container. Water equivalent of container is 0.50 kg and final temperature of mixture is 80oC. Find the final mass (approximate) of water in container. (Lvap=540 cal/g, swater=1calg−oC)