One end of a `2.35m` long and `2.0cm` radius aluminium rod `(K=235W.m^(-1)K^(-1))` is held at `20^(@)`C .The other end of the rod is in contact with a block of ice at its melting point.The rate in `kg.s^(-1)` at which ice melts is [Take latent heat of fusion for ice as `(1)/(3)times10^(6)``J.kg^(-1)`]

How many grams of ice at 0^@C can be melted by the addition of 500 J of heat ? (The molar heat of fusion for ice is 6.02 Kamal^(-1))

When 1 kg of ice at 0^(@)C melts to water at 0^(@)C , the resulting change in its entropy, taking latent heat of ice to be 80 cal//g is

A coil of water of resistance 50 Omega is embedded in a block of ice and a potential difference of 210 V is applied across it. The amount of ice which melts in 1 second is [ latent heat of fusion of ice = 80 cal g^(-1) ]

Calculate the entropy change in the melting of 1 kg of ice at 0^(@)C in SI units. Heat of funsion of ice = 80 cal g^(-1) .

When 1.5 kg of ice at 0^(@)C mixed with 2 kg of water at 70^(@)C in a container, the resulting temperature is 5^(@)C the heat of fusion of ice is ( s_("water") = 4186 j kg^(-1)K^(-1))

When 0.15 kg of ice at 0^(@)C is mixed with 0.30 kg of water at 50^(@)C in a container, the resulting temperature is 6.7^(@)C . Calculate the heat of fusion of ice. (s_(water) = 4186 J kg^(-1)K^(-1))

One end of a steel rod (K=46Js^(-1)m^(-1)C^(-1)) of length 1.0m is kept in ice at 0^(@)C and the other end is kept in boiling water at 100^(@)C . The area of cross section of the rod is 0.04cm^(2) . Assuming no heat loss to the atmosphere, find the mass of the ice melting per second. Latent heat of fusion of ice =3.36xx10^(5)Jkg^(-1) .

Calculate the power of an electric heater required to melt 1 kg of ice at 0^@ C in 30 s if the efficiency of heater is 40%. Take specific latent heat of ice = 336 J g^(-1)

Calculate the change in entropy for the fusion of 1 mol of ice. The melting point of ice is 300K and molar enthalpy of fustion for ice = 6.0 k J mol^(-1) .

A slab of stone of area of 0.36 m^(2) and thickness 0.1 m is exposed on the lower surface to steam at 100^(@)C . A block of ice at 0^(@)C rests on the upper surface of the slab. In one hour 4.8 kg of ice is melted. The thermal conductivity of slab is (Given latent heat of fusion of ice = 3.63 xx 10^(5) J kg^(-1) )