Four resistances carrying a current shown in Fig. `7.41` are immersed in a box containing ice at `0^(@)C`. How much ice must be put in the box every `10 min` to keep the average quantity of ice in the box constant? Latent heat of ice is `80 calg^(-1)`.

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

Consider the following arrangement of resistors . This arrangement is immersed in a box containting ice at 0^@)C . How much ice must be put in the box constant? (Latent heat of ice L = 80 cal//g ) .

A 10 H inductor carries a current of 20A . How much ice at 0^@C could melted by the energy stored in the magnetic field of the inductor? Latent heat of ice is 22.6xx10^I3J//kg

As shown in Fig. AB is rod of length 30 cm and area of cross section 1.0 cm^2 and thermal conductivity 336 SI units. The ends A and B are maintained at temperatures 20^@C and 40^@C , respectively .A point C of this rod is connected to a box D, containing ice at 0^@C through a highly conducting wire of negligible heat capacity. The rate at which ice melts in the box is (assume latent heat of fusion for ice L_f=80 cal//g )

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

One gram of ice at 0^(@)C is added to 5 gram of water at 10^(@)C . If the latent heat of ice be 80 cal/g, then the final temperature of the mixture is -

Rays from the sun ar focuseed by a lens of diameter 5 cm on to a block of ice and 10 g of ice is melted in 20 min. Therefore the heat from the sun reaching the earth per min per square centrimetre is (Latent heat of ice L = 80 cal g^(-1) )