The two opposite faces of a cubical piece of iron (thermal conductivity = 0.2 CGS units) are at `100^(@)C` and `0^(@)C` in ice. If the area of a surface is `4cm^(2)` , then the mass of ice melted in 10 minutes will b

The opposite faces of a cubical block of iron and cross-section 4 sq. cm are kept in contact with steam of melting ice. Calculate the amount of ice melted at the end of 10 minutes if K = 0.2 cal cm^-1s^-1C^-1 for iron.

The opposite faces of a cubical block of iron of cross-section 4 xx 10^(-4) "m"^(2) are kept in contact with steam and melting ice. Calculating the amount of ice melted at the end of 5 minutes. Given K=0.2 cal cm ""^(-1) "s"^(-1)°"C"^(-1) .

The opposite faces of a cubical block of iron of cross-section 4 square cm are kept in contact with steam and melting ice. Calculate the quantity of ice melted at the end of 10 minutes. k for iron = 0.2 CGS units.

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

A 3 cm cube of iron has one face at 100^(@)C and the other in a block of ice at 0^(@)C . If k of iron = 0.2 CGS units and L for ice is 80cal/gm, then the amount of ice that melts in 10 minutes is (assume steady state heat transfer)

A rectangular block of copper (K= 0. 9). of thickness 5cm and area of cross section 10 cm^(2) has one of its faces maintained at a constant temperature of 100^(@) C while the opposite face is in contact with ice at 0^(@)C . If there is no loss of heat, the amount of ice that melts in 10 minutes is