Steam at 100- C is passed into 1-1 kg of water contained in a calorimeter of water equivalent 0-02 kg at 15- C till the temperature of the calorimeter and its contents rises to 80- C- The mass of the steam condensed in kg isIIT JEE 1986A- 0-13B- 0-065C- 0-26D- 0-135

The correct option is D 0.135Heat required Q = (1.1 + 0.02) × 10^3 × 1 × (80 15) = 72800 cal. Therefore, mass of steam condensed (in kg) m = Q/L = 72800/ ...

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Standard X

Physics

Latent Heat

Steam at 100∘...

Question

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

(IIT JEE 1986)

0.13

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0.065

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0.26

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0.135

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Solution

The correct option is D 0.135
Heat required
$Q = (1.1 + 0.02) \times 10^{3} \times 1 \times (80 - 15) = 72800 c a l .$
Therefore, mass of steam condensed (in kg)
$m = \frac{Q}{L} = \frac{72800}{540} \times 10^{- 3} = 0.135 k g .$

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Steam at 100oC is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at 15oC till the temperature of the calorimeter and its contents rises to 80oC. The mass of the steam condensed in kg is (IIT JEE 1986)

The correct option is D 0.135Heat required Q=(1.1+0.02) × 103 × 1 × (80−15)=72800 cal. Therefore, mass of steam condensed (in kg) m=QL=72800540 × 10−3=0.135kg.

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

(IIT JEE 1986)

The correct option is D 0.135
Heat required
$Q = (1.1 + 0.02) \times 10^{3} \times 1 \times (80 - 15) = 72800 c a l .$
Therefore, mass of steam condensed (in kg)
$m = \frac{Q}{L} = \frac{72800}{540} \times 10^{- 3} = 0.135 k g .$