The only possibility of heat flow in a thermos flask is through its cork which is `75 cm^(2)` in area and 5 cm thick. Its thermal conductivity is `0.0075 cal//cm sec^(@)C`. How lolng will 500 g of ice at `0^(@)C` in thermos flask will is `40^(@)C` and latent heat of ise is `80 cal.//gram`.

The only possibility of heat flow in a thermos flask is through its cork which is 75cm^2 in area and 5 cm thick its thermal conductivity is 0.0075 cal//cm-s-^@C . The outside temperature is 40^@C and latent heat of ice is 80 cal//g . Time taken by 500 g of ice at 0^@C in the flask to melt into water at 0^@C is

The only possibility of heat flow in a thermos flask is through its cork which is 75cm^2 in area and 5 cm thick its thermal conductivity is 0.0075 cal//cm-s-^@C . The outside temperature is 40^@C and latent heat of ice is 80 cal//g . Time taken by 500 g of ice at 0^@C in the flask to melt into water at 0^@C is

The only possibility of heat flow in a thermos flask is through its cork which is 75cm^2 in area and 5 cm thick its thermal conductivity is 0.0075 cal//cm-s-^@C . The outside temperature is 40^@C and latent heat of ice is 80 cal//g . Time taken by 500 g of ice at 0^@C in the flask to melt into water at 0^@C is

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

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

How many calories of heat will be absorbed when 2kg of ice at 0^(@)C melts ? ( Specific latent heat of fusion of ice = 80 cal //g )

What amount of ice will remains when 52 g ice is added to 100 g of water at 40^(@)C ? Specific heat of water is 1 cal/g and latent heat of fusion of ice is 80 cal/g.

Calculate the heat gained by 50 g of ice at 0^(@)C to change into water of the same temperature. (Latent heat of fusion of ice is 80 cal/g)