Steam at $100$ $^{o}$ C is passed into a calorimeter of water equivalent $10$ g containing $94$ cc of H $_{2}$ O and $10$ g of ice at $0^{o}$ C. The temperature of the calorimeter and contents rise by $5^{o}$ C. The amount of steam passed is

1 g

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2 g

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3 g

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4 g

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Solution

The correct option is A 2 g
Let $m_{S}$ be the amount of steam in grams.
Then $m_{S} L + m_{S} C_{w} (100 - 5)$
$= W_{c a l} (5 - 0) + 10 \times 80 + (104) (5 - 0)$
$m_{5} (540) + m_{S} (95) = 10 (5) + 800 + 420$
$653 m_{S} = 1270$ or $m_{S} = 2 g$

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(IIT JEE 1986)

Q. Steam at

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is passed into

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Steam at $100^{o} C$ is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at $15^{o} C$ till the temperature of the calorimeter and its contents rises to $80^{o} C .$ The mass of the steam condensed in kg is
(Latent heat of vapor = $540 c a l / g$ )