1) The walls of a house are 24 cm thick and have an average thermal conductivity four times that of glass wool. The walls' surface area is 800000 cm2 and their inside surface is at 21°C, while their outside surface is at 5°C. (Thermal conductivity of glass wool is 0.042 W/m°C). Calculate, (i) The rate of heat conduction through house walls: (ii) Amount of heat in Joules, conducted in 5 min:

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Chapter1: Units, Trigonometry. And Vectors
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1) The walls of a house are 24 cm thick and have
an average thermal conductivity four times that
of glass wool. The walls' surface area is
800000 cm? and their inside surface is at 21°C,
while their outside surface is at 5°C. (Thermal
conductivity of glass wool is 0.042 W/m°C).
Calculate,
(i) The rate of heat conduction through house
walls:
(ii) Amount of heat in Joules, conducted in 5
min:
2) A spherical infrared heater of radius 6.3 cm
and an emissivity of 0.79 radiates 0.48 kW of
power. Given, Stefan's constant = 5.67x108 Wm¯
2K4. Calculate,
Temperature of the heater in Celsius:
Transcribed Image Text:1) The walls of a house are 24 cm thick and have an average thermal conductivity four times that of glass wool. The walls' surface area is 800000 cm? and their inside surface is at 21°C, while their outside surface is at 5°C. (Thermal conductivity of glass wool is 0.042 W/m°C). Calculate, (i) The rate of heat conduction through house walls: (ii) Amount of heat in Joules, conducted in 5 min: 2) A spherical infrared heater of radius 6.3 cm and an emissivity of 0.79 radiates 0.48 kW of power. Given, Stefan's constant = 5.67x108 Wm¯ 2K4. Calculate, Temperature of the heater in Celsius:
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