A brass rod with a length of 20.0 cm is placed side by side with an aluminum rod with a length of 20.0 cm and this system is placed between a hot temperature of 150$^{\circ}$C and a cold temperature of -10.0$^{\circ}$C.The coefficient of thermal conductivities of the brass and the aluminum are 100 W/m$^{\circ}$C and 230 W/m C respectively.The rods have the same cross sectional area of 20.0 cm².What is the heat flow from the hot temperature to the cold temperature?

Question

Quizplus · Accepted Answer

Using Fourier's law of heat conduction, the heat flow can be calculated as:

Q = k*A*(delta T)/d

where Q is the heat flow, k is the thermal conductivity, A is the cross sectional area, delta T is the temperature difference, and d is the length of the rods.

For the brass rod:

Q_b = (100 W/m C)*(0.002 m^2)*(150-(-10) C)/(0.2 m)
    = 500 W

For the aluminum rod:

Q_a = (230 W/m C)*(0.002 m^2)*(150-(-10) C)/(0.2 m)
    = 1150 W

The heat flow through the two rods in series is the same, and is given by:

Q_tot = Q_b = Q_a

Q_tot = 500 W

Therefore, the best choice is B, with a heat flow of 528 W.

A Brass Rod with a Length of 20 ∘^{\circ}∘ C and a Cold Temperature of -10

A Brass Rod with a Length of 20 $^{\circ}$ C and a Cold Temperature of -10