When 1g of water at `0^@C` and `1xx10^5(N)/(m^2)` pressure is converted into ice of volume `1.091cm^3`. The external work done will e

When 1g of water at 0^(@)C and 1 xx 10^(4)Nm^(-2) pressure is converted into ice of volume 1.091 cm^(3) , find the work done by water? (rho_(w) = 1 gm//cm^(3))

When 1g of water at 0^(@)C and 1 xx 10^(4)Nm^(-2) pressure is converted into ice of volume 1.091 cm^(3) , find the work done by water? (rho_(w) = 1 gm//cm^(3))

How much work is done agaist uniform pressure when 1g of water at 100^(@)C is converted into steam? Express your result in calories (J=4.2 joules per calorie volume of 1kg of steam at 100^(@)C=1650xx10^(-3)m^(3) and 1 atmospheric pressure =10^(5)Nm^(-2) )

At 100^(@)C the volume of 1 kg of water is 10^(-3) m^(3) and volume of 1 kg of steam at normal pressure is 1.671 m^(3) . The latent heat of steam is 2.3 xx 10^(6) J//kg and the normal pressure is 5xx10^(5) N//m^(2) . If 5 kg of water at 100^(@)C is converted into steam, the increase in the internal energy of water in this process will be

A sample of 0.1 g of water at 100^(@)C and normal pressure (1.013 xx 10^(5) N cdot m^(-2)) requires 54 cal of heat energy to convert to steam at 100^(@)C . If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample is

At 100^(@)C the volume of 1 kg of water is 10^(-3) m^(3) and volume of 1 kg of steam at normal pressure is 1.671 m^(3) . The latent heat of steam is 2.3 xx 10^(6) J//kg and the normal pressure is 10^(5) N//m^(2) . If 5 kg of water at 100^(@)C is converted into steam, the increase in the internal energy of water in this process will be

Bulk modulus of water is 2 xx 10^9 N m^-2 , The pressure required to increase the volume of water by 0.1% in Nm^-2 is

1 g of water (volume 1 cm^(3) becomes 1671 cm^(3) of steam when boiled at a pressure of 1atm. The latent heat of vaporisation is 540 cal/g, then the external work done is (1 atm = 1.013 xx 10^(5) N/ m^(2))

A Carnot refrigerator absorbs heat from water at 0^(@)C and gives it to room at 27^(@)C , when it converts 2 kg of water at 0^(@)C into ice at 0^(@)C , the work done is (latent heat of fusions of ice= (333xx10^(3) Jkg^(-1))

When water is boiled at 2 atm pressure the latent heat of vapourization is 2.2xx10^(6) J//kg and the boiling point is 120^(@)C At 2 atm pressure 1 kg of water has a volume of 10^(-3)m^(-3) and 1 kg of steam has volume of 0.824m^(3) . The increase in internal energy of 1 kg of water when it is converted into steam at 2 atm pressure and 120^(@)C is [ 1 atm pressure =1.013xx10^(5)N//m^(2) ]