The concrete slab of a basement is 11 m long, 8 m wide, and 0.20 m thick. During the winter, temperatures are nominally 17 C and 10 C at the top and bottom surfaces, respectively. If the concrete has a thermal conductivity of 1.4 W/m K, what is the rate of heat loss through the slab

Q = - 4312 W = - 4.312 KW

Explanation:

The rate of heat of the concrete slab can be calculated through Fourier's Law of heat conduction. The formula of the Fourier's Law of heat conduction is as follows:

Q = - kA dt/dx

Integrating from one side of the slab to other along the thickness dimension, we get:

Q = - kA(T₂ - T₁)/L

Q = kA(T₁ - T₂)/t

where,

Q = Rate of Heat Loss = ?

k = thermal conductivity = 1.4 W/m.k

A = Surface Area = (11 m)(8 m) = 88 m²

T₁ = Temperature of Bottom Surface = 10°C

T₂ = Temperature of Top Surface = 17° C

t = Thickness of Slab = 0.2 m

Therefore,

Q = (1.4 W/m.k)(88 m²)(10°C - 17°C)/0.2 m

Q = - 4312 W = - 4.312 KW

Here, negative sign shows the loss of heat.

the total distance traveled by the car is

30km + 49km = 79 km .