Question

A room in a house has a floor made entirely of concrete which is 200 mm thick. The lower surface of the concrete in contact with the ground has a temperature of 10°C and the upper surface, in contact with the living area has a temperature of 15°C. The floor is square and has sides 10m x 10m

a) Calculate the rate at which thermal energy ia conducted through the concrete. Assume the thermal conductivity of concrete is 0.750W/mK.

b) The temperature of thr carpet/concrete interface. The rate of conduction for both conductors are the same and the thermal conductivity of the carpet is 0.06W/mk.

c) The rate at which thermal energy is conducted through the two layers.

Answer 1

Answer on Question #80115, Physics / Molecular Physics | Thermodynamics

A room in a house has a floor made entirely of concrete which is 200 mm thick. The lower surface of the concrete in contact with the ground has a temperature of 10°C and the upper surface, in contact with the living area has a temperature of 15°C. The floor is square and has sides 10m x 10m

a) Calculate the rate at which thermal energy is conducted through the concrete. Assume the thermal conductivity of concrete is 0.750W/mK.

b) The temperature of the carpet/concrete interface. The rate of conduction for both conductors are the same and the thermal conductivity of the carpet is 0.06W/mK.

c) The rate at which thermal energy is conducted through the two layers.

Answer:

a)

$$\frac {\Delta Q}{\Delta t} = \frac {k A}{l} (\theta_ {1} - \theta_ {2})$$ $$\frac {\Delta Q}{\Delta t} = \frac {0.75 \times 100}{0.2} (15 - 10) = 1875\,W = 1.875\,kW$$

b)

$$\frac {\Delta Q}{\Delta t} = \frac {k _ {1} A}{l _ {1}} (\theta_ {1} - \theta_ {3})$$ $$\frac {\Delta Q}{\Delta t} = \frac {0.06 \times 100}{0.015} (15 - \theta_ {3}) = 400 (15 - \theta_ {3})$$ $$\frac {\Delta Q}{\Delta t} = \frac {k _ {2} A}{l _ {2}} (\theta_ {3} - \theta_ {2})$$ $$\frac {\Delta Q}{\Delta t} = \frac {0.75 \times 100}{0.2} (\theta_ {3} - 10) = 375 (\theta_ {3} - 10)$$ $$400 (15 - \theta_ {3}) = 375 (\theta_ {3} - 10)$$ $$\theta_ {3} = 12.58\,{}^{\circ}\mathrm{C}$$

c)

$$\frac {\Delta Q}{\Delta t} = 400(15 - 12.58) = 968\,W = 0.968\,kw$$

Answer: a) 1.875 kW; b) 12.58 °C; c) 0.968 kw

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