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1 of a steam at 100^@C melts how much ic...

1 of a steam at `100^@C` melts how much ice at `0^@C`? (`Latent heat of ice=80 cal/g` and `latent heat of steam =540 cal/g`)

A

1 g

B

2 g

C

4 g

D

8 g

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How much steam at 100^@C will just melt 1600 g of ice at -8^@C ? (Specific heat of ice= 0.5 (cal)/g^@C , specific heat of water= 1/cal/g^@C , latent of fusion of ice = 80 cal/g , and latent heat of vaporisation of water= 540 cal/g )

M' kg of water at 't' ^@C is divided into two parts so that one part of mass 'm' kg when converted into ice 0^@C would release enough heat to vapourise the other part, then m/M is equal to [Specific heat of water = 1 calg^-1^@C^-1 Latent heat of fusion of ice= 80 calg^-1 Latent heat of steam = 540 cal g^-1 ]

If 80 g steam of temperature 97^@C is released on an ice slab of temperature 0^@C ,how much ice will melt?How much energy will be transferred to water? Given : specific heat capacity of water, c = 1 cal/g °C Latent heat of melting of ice = 80 cal/g Latent heat of vaporization of water = 540 cal/g

If 80 g steam of temperature 87^@C is released on an ice slab of temperature O^@C ,how much ice will melt?How much energy will be transferred to water? Given : specific heat capacity of water, c = 1 cal/g °C Latent heat of melting of ice = 80 cal/g Latent heat of vaporization of water = 540 cal/g

Latent Heat of fusion of ice is

Steam 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 )

If the mass of steam is 50g , how much ice will melt ?

Complete the analogy: Latent heat of fusion of ice : 80 cal/g :: steam:

60 g of ice at 0^@C is mixed with 60 g of steam at 100^@C . At thermal equilibrium, the mixture contains (Latent heat of steam and ice are 540 g^-1 and 80 cal g^-1 respectively, specific heat of water = 1cal g^-1^@C^-1 )

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