Answer:
answer a) 2
Step-by-step 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