A refrigerator door is 150 cm high, 80 cm wide, and 6 cm thick. If the coefficient of conductivity is 0.0005 `cal//cm s^@C` and the inner and outer surfaces are at `0^@C` and `30^@C`, respectively, what is the heat loss per minute through the door, in caloris?

A refrigerator door is 150 cm high, 80 cm wide and 6 cm thick. If the coefficient of conductiviy is 0.0005 (cal)/(cm s^@C) and the inner and outer surfaces are at O^@C and 30^@C respectively, then what is the heat loss (in calories) per minute through the door?

The door of a refrigerator is 150 cm. high, 80 cm wide and 6 cm thick. If the coefficient of thermal conductivity is 5 xx 10^-4 cal//cm . s. ^@C and the inner and outer surfaces are at 0^@C and 30^@C respectively. Calculate heat loss through the door in 1 minute.

A metallic cylindrical shell of length 50 cm has inner radius of 3 cm and outer radius of 6 cm. If the inner and outer surfaces of the cylinder are maintained at 0^@C and 80^@C , respectively, then find the rate of flow of heat from the outer surface to the inner surface. Take, thermal conductivity of metal = 69.3 Wm^(-1) k^(-1)

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

If the coefficient of conductivity of aluminium is 0.5 cal cm^(-1) s^(-1).^(@)C^(-1) , then the other to conductor 10 cal s^(-1) cm^(-2) in the steady state, the temperature gradient in aluminium must be

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 .

Three rods of copper, brass and steel are welded together to form a Y -shaped structure. The cross-sectional area of each rod is 4 cm^2 . The end of copper rod is maintained at 100^@C and the ends of the brass and steel rods at 80^@C and 60^@C respectively. Assume that there is no loss of heat from the surfaces of the rods. The lengths of rods are : copper 46 cm, brass 13 cm and steel 12 cm. (a) What is the temperature of the junction point? (b) What is the heat current in the copper rod? K(copper) = 0.92, K (steel) = 0.12 and K(brass) = 0.26 cal//cm-s^@C .