Steam at 100°C is passed into 100 g of i...

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) `]

`80g`

`100g`

`90g`

`75g`

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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) .

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.

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)

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

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

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