If latent heat of fusion of ice is 80 ca...

If latent heat of fusion of ice is 80 cals per g at `0^(@),` calculate molal depression constant for water.

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The correct Answer is:

"1.86 K kg mol"^(-1)`

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A heater melts 0^@C ice in a bucket completely into water in 6 minutes and then evaporates all that water into steam in 47 minutes 30 sec . If latent heat of fusion of ice is 80 cal//gram , latent heat of steam will be (specific heat of water is 1 cal//gam-^@C )

Specific latent of fusion of ice is 80 cal/g. Explain this statement ?

A vessel with 100 g of water at a temperature of 0^(@)C is suspended in the middle of a room. In 15 minutes the temperature of the water rises to 1.8^(@)C . When ice equal in weight of the water is placed in the same vessel, it melts during 10 hours. using appropriate appropriate approximations, estimate the latent heat of fusio of ice in cal/g. if the known alue of latent heat of fusion of ice is 80 cal/g, obtain the difference in the two values in cal/g and report this as your answer.

1 mole of ice at 0^(@)C and 4.6 mm Hg pressure is converted to water vapour at a constant temperature and pressure. Find Delta H and Delta E if the latent heat of fusion of ice is 80 cal//g and latent heat of vaporisation of liquid water at 0^(@)C is 596 cal//g and the volume of ice in comparison to that of water (vapour) in neglected.

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