Carly places one end of a steel bar in a Bunsen flame and the other end in an ice cube.

By what factor is the rate of heat flow changed when the bar's cross-sectional area is doubled?

a. 2b. 1/2c. 4.0d. 1/4

answer a) 2

Explanation:

Assuming stationary state, following Fourier's law:

Q = A*k * dT/dL

where Q = heat flow, A = cross sectional area, dT/dL = temperature gradient along the bar

if the cross sectional area is doubled, then the gradient is the same (since the heat sources do not change in temperature or position, and the length of the bar is the same). Since the gradient is same, the temperature is the same under stationary conditions, then we can assume k remains constant in the cross section. Therefore

Q₁ = A₁*k * dT/dL

Q₂ = A₂*k * dT/dL

dividing both equations

Q₂ / Q₁ = A₂/A₁ = 2

then the correct answer is a)

Note:

Since the cross sectional area is doubled, then heat loss to the surroundings will be

Q loss = h * A exposed * ΔT

then

Q loss₂ / Q loss ₁ = A exposed ₂ / A exposed ₁

for a circular cross section or a squared cross section

A exposed ₂ / A exposed ₁ = √2

then

Q loss₂ / Q loss ₁ = √2

therefore we did not take into account the increase in heat loss due to the increased in exposed area to the environment