Home
Class 12
PHYSICS
Steam at 100^(@)C is passed into 1 kg of...

Steam at `100^(@)C` is passed into 1 kg of water contained in a calorimeter at `9^(@)C` till the temperature of water and calorimeter is increased to `90^(@)C`. The mass of the steam condensed is nearly
(water equivalent of calorimeter =0.1 kg, specific heat of water `=1" cal g"^(-1)""^(@)C^(-1)` and latent heat of vaporisation `=540" cal g"^(-1)`)

A

81 g

B

162 g

C

243 g

D

486 g

Text Solution

Verified by Experts

The correct Answer is:
B
Let mass of the steam condensed is x. Heat released = Heat gained by water
`rArr x xx 540 +x xx 1xx(100-90)`
`=1xx1xx(90-9)+0.1xx1xx(90-9)`
`rArr 540 x+10x=81+8.1`
`rArr" "x=(891)/(550) =0.162 kg, x=162 g`
Promotional Banner

Similar Questions

Explore conceptually related problems

A steam at 100^@C is passed into 1 kg of water contained in a calorimeter of water equivalent 0.2 kg at 9^@C till the temperature of the calorimeter and water in it is increased to 90^@C . The mass of steam condensed in kg is nearly (specific heat of water = 1 cal//g^@C ,latent heat of vaporisation= 540 cal // g)

Steam at 100^(@)C is passed into 22 grams of water at 20^(@)C . When resultant temperature is 90^(@)C , then weight of the water present is

Steam at 100^(@)C is passed into 20 g of water at 10^(@)C . When water axquires a temperature of 80^(@)C , the mass of water present will be [Take specific heat of water 1"cal"g^(-1)c^(-1) latent heat of steam = 540 cal//g^(-1) ]

1 g of steam at 100^(@)C is passed into a calorimeter of water equivalent 10g containing water of mass 90 g. Find the resultant temperature of the mixture if the initial temperature of water is 27^(@)C .