Steam at 120^(@)C` is continuously passed through a `50-cm` long rubber tube of inner and outer ratii `1.0cm` and `1.2cm` . The room temperature is `30^(@)C` . Calculate the rate of heat flow through the walls of the tube. Thermal conductivity of rubber `= 0.15Js^(-1)m^(-1)`(@)C^(-1)` .

A room has s window fitted with a single 0.1mxx2.0m glass of thickness 2mm. (a) Calculate the rate of heat flow through the clossed window when the temperature inside the room is 32^(@)C and that outside is 40^(@)C . (b) The glass is now replaced by two glasspance, each having a thickness of 1mm and separated by a distance of 1mm. Calculalte the rate of heat flow under the same condition of temperature. Thermal conductivity of window glass =1.0Js^(-1)m^(-1) ^(@)C^(-1) and that of air =0.025Js^(-1)m^(-1) ^(@)C^(-1) .

A room has s window fitted with a single 1.0mxx2.0m glass of thickness 2mm. (a) Calculate the rate of heat flow through the closed window when the temperature inside the room is 32^(@)C and that outside is 40^(@)C . (b) The glass is now replaced by two glasspanes, each having a thickness of 1mm and separated by a distance of 1mm. Calculate the rate of heat flow under the same condition of temperature. Thermal conductivity of window glass =1.0Js^(-1)m^(-1)C^(-1) and that of air =0.025Js^(-1)m^(-1)C^(-1) .

A pitcher with 1-mm thick porous walls contain 10kg of water. ,Water comes to its outer surface and evaporates at the rate of 0.1gs^(-1) . The surface area of the pitcher (one side) =200cm^(2) . The room temperature =42^(@)C , latent heat of vaporization =2.27xx10^(6)Jkg^(-1) , and the thermal conductivity of the porous walls =0.80js^(-1)m^(-1) ^(@)C^(-1) . Calculate the temperature of water in the pitcher when it attains a constant value.

A rod of 1m length and area of cross-section 1cm^(2) is connected across two heat reservoirs at temperature 100^(@)C and 0^(@)C as shown. The heat flow per second through the rod in steady state will be [Thermal conductivity of material of rod =0.09kilocalm^(-1)s^(-1)('^(@)C) ]

What is the temperature of the steel-copper junction in the steady state system show in Fig. 7(e).17 ? Length of steel rod = 15.0 cm, length of the copper rod = 10.0 cm, temperature of the furnace =300^(@)C , temperature of other end 0^(@)C . The are of cross-section of the steel rod is twice that of the copper rod. (Thermal conductivity of steel =50.2 js^(-1) m^(-1) K^(-1) and of copper =3895 js^(-1) m^(-1) K^(-1) ).

What is the temperature of the steel-copper junction in the steady state system show in Fig. 7(e).17 ? Length of steel rod = 15.0 cm, length of the copper rod = 10.0 cm, temperature of the furnace =300^(@)C , temperature of other end 0^(@)C . The are of cross-section of the steel rod is twice that of the copper rod. (Thermal conductivity of steel =50.2 js^(-1) m^(-1) K^(-1) and of copper =3895 js^(-1) m^(-1) K^(-1) ).

A pitcher with 1-mm thick porous walls contain 10kg of water. ,Water comes to its outer surface and evaporates at the rate of 0.1gs^(-1) . The surface area of the pitcher (one side) =200cm^(2) . The room temperature =42^(@)C , latent heat of vaporization =2.27xx10^(6)Jkg^(-1) , and the thermal conductivity of the porous walls =0.80js^(-1)m^(-1)C^(-1) . Calculate the temperature of water in the pitcher when it attains a constant value.

A pitcher with 1-mm thick porous walls contain 10kg of water. ,Water comes to its outer surface and evaporates at the rate of 0.1gs^(-1) . The surface area of the pitcher (one side) =200cm^(2) . The room temperature =42^(@)C , latent heat of vaporization =2.27xx10^(6)Jkg^(-1) , and the thermal conductivity of the porous walls =0.80js^(-1)m^(-1)C^(-1) . Calculate the temperature of water in the pitcher when it attains a constant value.

Calculate the quantity of heat conducted through 2m^(2) of a brick wall 12cm thick in 1 hour if the temperature on one side is 8^(@)C and on the other side is 28^(@)C . (Thermal conductivity of brick =0.13Wm^(-1)K^(-1) )