The thickness of a brick wall is 0.25m. The temperatures inside and outside or `24^(@)` C and `42^(@)`C. Given Thermal conductivity of brick = 0.15`Wm^(-1)K^(-1)` , then calculate the total amount of heat conducted in a time interval of 30 minutes due to a 25`m^2` wall.

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) )

The thermal conductivity of brick is 1.7 W m^(-1) K^(-1) , and that of cement is 2.9 W m^(-1) K^(-1) .What thickness of cement will have same insulation as the brick of thickness 20 cm .

The thermal conductivity of brick is 1.7 W m^(-1) K^(-1) , and that of cement is 2.9 W m^(-1) K^(-1) .What thickness of cement will have same insulation as the brick of thickness 20 cm .

The thermal conductivity of brick is 1.7 W m^(-1) K^(-1) , and that of cement is 2.9 W m^(-1) K^(-1) .What thickness of cement will have same insulation as the brick of thickness 20 cm .

A wall of a room has dimensions of 6m xx 4m . It is 60 cm thick. Its thermal conductivity is is 2. If the temperature inside and outside is 20^@C and -5^@C respectively the amount of heat conducted every hour is about

The conductivity of 0.25 M solution of KCI at 300 K is 0.0275 cm^(-1) calculate molar conductivity

The conductivity of 0.25 M solution of KCI at 300 K is 0.0275 cm^(-1) calculate molar conductivity

A copper slab is 2 mm thick. It is protected by a 2 mm layer of stainless steel on both sides. The temperature on one side of this composite slab is 400^(@)C and 200^(@)C on the other side. Value of thermal conductivities are- k_(cu)=400 Wm^(-1)K^(-1) and k_(s)=16 Wm^(-1)K^(-1) (a) Just by knowing that thermal conductivity of steel is much less than copper, find (approximately) the temperature of the copper slab. (b) Plot the variation of temperature across the thickness of the composite wall.

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.