Steam at 100 ° C is passed into 1.1 textkg of water contained in a calorimeter of water equivalent 0.02 textkg at 15 ° C , till the temperature of the calorimeter and its contents rises to 80 text °C text. The mass of steam condensed (in kg ) is (. Take latent heat of steam = 540 cal g- 1) :

Question

Steam at 100 ° C is passed into 1.1 textkg of water contained in a calorimeter of water equivalent 0.02 textkg at 15 ° C , till the temperature of the calorimeter and its contents rises to 80 text °C text. The mass of steam condensed (in kg ) is (. Take latent heat of steam = 540 cal g- 1) : (A) 0.130 (B) 0.065 (C) 0.260 (D) 0.135

Tardigrade Admin · Accepted Answer

Heat release by steam = Heat gained by calorimeter and its contents Solution

1.12× (80 - 15)=m× 540+m× 20 m=0.130 kg

$0.130$

$0.065$

$0.260$

$0.135$

Heat release by steam = Heat gained by calorimeter and its contents

$1.12 \times (80 - 15) = m \times 540 + m \times 20$
$m = 0.130 k g$

Q. Steam at 100∘C is passed into 1.1kg of water contained in a calorimeter of water equivalent 0.02kg at 15∘C , till the temperature of the calorimeter and its contents rises to 80 °C. The mass of steam condensed (in kg ) is ( Take latent heat of steam =540calg−1) :

0.130

0.065

0.260

0.135

Heat release by steam = Heat gained by calorimeter and its contents 1.12×(80−15)=m×540+m×20 m=0.130kg

$0.130$

$0.065$

$0.260$

$0.135$

Heat release by steam = Heat gained by calorimeter and its contents

$1.12 \times (80 - 15) = m \times 540 + m \times 20$
$m = 0.130 k g$