Question

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

Answer 1

Answer:

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