Calculate the velocity with which a piece of ice at - `5^(@)C` be thrown against a wall so that its entire mass melts upon contact. Latent heat of ice = 80 cal `g^(-1)`, specific heat of ice = 0.5 cal `g^(-1).^(@)C^(-1)`. And `J=4.2 xx 10^(7)` erg `cal^(-1)`.

The entropy change in melting 1g of ice at 0^@C . (Latent heat of ice is 80 cal g^(-1) )

75 g of water at 100^@ C is added to 20 g of ice at -15^@C . Calculate the resulting temperature, give that latent heat of ice= 80 cal g^(-1)C^(-1) and specific heat of ice=0.5 cal g^(-1) C^(-1) .

The water of mass 75 g at 100^(@)C is added to ice of mass 20 g at -15^(@)C . What is the resulting temperature. Latent heat of ice = 80 cal//g and specific heat of ice = 0.5 .

Calculate the amount of heat required to convert 10g of ice at - 10^@C to water at 5^@C (latent heat of fusion of ice = 80 cal g^(-1) , specific heat of ice = 0.5).

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 )

What is the amount of work done to convert 40 g of ice at -10^(@)C into steam at 100^(@)C ? Specific heat of ice = 0.5 cal cdot g^(-1) cdot ^(@)C^(-1) , latent heat of fusion of ice =80 cal cdot g^(-1) , latent heat of steam = 540 cal cdot g^(-1), J = 4.2 xx 10^(7) erg cdot cal^(-1) .

An iron ball of mass 0.2 kg is heated to 10^(@)C and put into a block of ice at 0^(@)C . 25 g of ice melts. If the latent heat of fusion of ice is 80 cal g^(-1) , then the specific heat of iron in cal g^(-1^(@))C is

An iron ball of mass 0.2 kg is heated to 10^(@)C and put into a block of ice at 0^(@)C . 25 g of ice melts. If the latent heat of fusion of ice is 80 cal g^(-1) , then the specific heat of iron in cal g^(-1^(@))C is