One mole of ice is melted at `0^(@)C` and then is heated to `100^(@)C`. What is the difference in entropies of the steam and ice? The heats of vaporisation and fusion are `540 cal g^(-1) and 80 cal g^(-1)` respectively. Use the average heat capacity of liquid water as 1cal `g^(-1) degree^(-1)`

A mole of steam is condensed at 100^(@) C, the water is cooled to 0^(@) C and frozen to i.e . What is the difference in entropies of the stem and ice? The heat of vaporization and fusion are 540 cal" "gm^(-1) and 80 cal" "gm ^(-1) respectively . Use the average heat capacity of liquild water as 1 cal" "gm^(-1) degree^(-1) .

How much heat is required to change 10g ice at 10^(@)C to steam at 100^(@)C ? Latent heat of fusion and vaporisation for H_(2)O are 80 cl g^(-1) and 540 cal g^(-1) , respectively. Specific heat of water is 1cal g^(-1) .

Steam at 100°C is passed into 100 g of ice at 0°C . Finally when 80g ice melts the mass of water present will be [Latent heat of fusion and vaporization are 80 cal g^(-1) and 540 cal g^(-1) respectively, specific heat of water = 1 cal g^(-1) °C^(-1) ]

How much heat is required to change 5 gram ice (0^@C) to steam at 100^@C ? Latent heat of fusion and vaporization for water are 80 cal/g and 540 cal/g respectively . Specific heat of water is 1 cal/ g/k.

Calculate the amount of heat required in calorie to change 1 g of ice at -10^(@)C to steam at 120^(@)C . The entire process is carried out at atmospheric pressure. Specific heat of ice and water are 0.5 cal g^(-1) .^(@)C^(-1) and 1.0 cal g^(-1) .^(@)C^(-1) respectively. Latent heat of fusion of ice and vaporization of water are 80 cal g^(-1) and 540 cal g^(-1) respectively. Assume steam to be an ideal gas with its molecules having 6 degrees of freedom. Gas constant R = 2 cal mol^(-1) K^(-1) .

What is the amount of heat required (in calories) to convert 10 g of ice at -10^(@)C into steam at 100^(@)C ? Given that latent heat of vaporization of water is "540 cal g"^(-1) , latent heat of fusion of ice is "80 cal g"^(-1) , the specific heat capacity of water and ice are "1 cal g"^(-1).^(@)C^(-1) and "0.5 cal g"^(-1).^(@)C^(-1) respectively.

The latent heat of vaporisation of water at 100 ^(@)C is 540 cal g^(-1) . Calculate the entropy increase when one mole of water at 100^(@)C is evaporated

Latent heat of vaporisation of water is 540cal g^(-1) at 100^(@)C calculate the entropy change when 1000 g water is converted to steam at 100^(@)C