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 kilograms is

The correct option is A 0.130 Heat required, Q = (1.1 + 0.02) × 10^3× 1× (80 15) = 72800 cal Let m gram of steam is condensed, then heat loss ...

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Byju's Answer

Standard XII

Physics

Calorimetry

Steam at 10...

Question

Steam at $100^{\circ} C$ is passed into $1.1 k g$ of water contained in a calorimeter of water equivalent $0.02 k g$ at $15^{\circ} C$ till the temperature of the calorimeter and its contents rises to $80^{\circ} C$ . The mass of the steam condensed in kilograms is

0.130

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0.065

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0.260

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0.135

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Solution

The correct option is A $0.130$
Heat required,
$Q = (1.1 + 0.02) \times 10^{3} \times 1 \times (80 - 15) = 72800 c a l$
Let $m$ gram of steam is condensed, then heat loss
$= m \times 540 + m \times 1 \times 20$
$= 560 m$
Heat loss $=$ Heat gain
$\Rightarrow 560 m = 72800$
$\Rightarrow m = 130 g = 0.130 k g$ .

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Similar questions

Q. Steam at

100^{\circ} C

is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at

15^{\circ} C

till the temperature of the calorimeter and its contents rise at

80^{\circ} C

. The mass of the stem condensed in kilogram is

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 = $540 c a l / g$ )

Q. Some steam at

100^{\circ} C

is passed into

1.1 k g

of water contained in a calorimeter of water equivalent

0.02 k g

15^{\circ} C

so that the temperature of the calorimeter and its contents rises to

80^{\circ} C

. What is the mass of steam condensing (in

k g

)

Specific heat of water = $1 k c a l / k g^{\circ} C$ , Latent heat of Vaporisation = $540 k c a l / k g$

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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 kilograms is

The correct option is A 0.130Heat required,Q=(1.1+0.02)×103×1×(80−15)=72800 calLet m gram of steam is condensed, then heat loss=m×540+m×1×20=560 mHeat loss = Heat gain⇒560 m=72800⇒m=130 g=0.130 kg.

Steam at $100^{\circ} C$ is passed into $1.1 k g$ of water contained in a calorimeter of water equivalent $0.02 k g$ at $15^{\circ} C$ till the temperature of the calorimeter and its contents rises to $80^{\circ} C$ . The mass of the steam condensed in kilograms is