A calorimeter of water equivalent 15 g contains 165 g of water at 25∘C. Steam at 100∘C is passed through the water for some time. The temperature is increased to 30∘C and the mass of the calorimeter and its contents is increased by 1.5 g. Calculate the specific latent heat of vaporization of water. The heat capacity of water is 1 cal g−1 ∘C−1
A calorimeter of water equivalent 15 g contains 165 g of water at 25∘C. Steam at 100∘C is passed through the water for some time. The temperature is increased to 30∘C and the mass of the calorimeter and its contents is increased by 1.5 g. Calculate the specific latent heat of vaporization of water. The heat capacity of water is 1 cal g−1 ∘C−1
The correct option is B 530 cal g−1
Let L be the specific latent heat of vaporization of water. The mass of the steam condensed is 1.5 g.
Heat lost in condensation of steam is Q1=(1.5g)L
The condensed water cools from 100∘C to 30∘C.
Heat lost in this process is -
Q2=(1.5 g)(1 cal g−1 ∘C−1)(70∘C)=105 cal
Heat supplied to the calorimeter and to the cold water during the rise in temperature from 25∘C to 30∘C is -
Q3=(15 g+165 g)(1 cal g−1 ∘C−1)(5∘C)=900 cal
If no heat is lost to the surrounding,
(1.5 g)L+105 cal=900 cal
or L=530 cal g−1
530 cal g−1
Let L be the specific latent heat of vaporization of water. The mass of the steam condensed is 1.5 g.
Heat lost in condensation of steam is Q1=(1.5g)L
The condensed water cools from 100∘C to 30∘C.
Heat lost in this process is -
Q2=(1.5 g)(1 cal g−1 ∘C−1)(70∘C)=105 cal
Heat supplied to the calorimeter and to the cold water during the rise in temperature from 25∘C to 30∘C is -
Q3=(15 g+165 g)(1 cal g−1 ∘C−1)(5∘C)=900 cal
If no heat is lost to the surrounding,
(1.5 g)L+105 cal=900 cal
or L=530 cal g−1
