Four rods with different radii $r$ and length $l$ are used to connect two heat reservoirs at different temperature. Which one will conduct most heat ?

Question

A 200 g sample of water at 80°C is mixed with 100 g of water at 20°C. Assuming no heat loss to the surroundings, what is the final temperature of the mixture?

Answer 1

To determine which rod will conduct the most heat, we must consider the formula for heat conduction:

Q = \dfrac{k \cdot A \cdot \Delta T \cdot t}{l}

where:

Q = Heat transfer
k = Thermal conductivity
A = Cross-sectional area of the rod
\Delta T = Temperature difference between the ends
t = Time during which heat transfer occurs
l = Length of the rod

The cross-sectional area A of a rod is given by \pi r^2.

Therefore, the heat transfer equation becomes:

Q \propto \dfrac{\pi r^2 \cdot \Delta T \cdot t}{l}

This shows that the heat conducted is directly proportional to the square of the radius and inversely proportional to the length of the rod.

Let's compare the rods:

r = 1 \, \text{cm}, \, l = 1 \, \text{m}
r = 1 \, \text{cm}, \, l = \frac{1}{2} \, \text{m}
r = 2 \, \text{cm}, \, l = 2 \, \text{m}
r = 2 \, \text{cm}, \, l = \frac{1}{2} \, \text{m}

For each rod, the expression for Q becomes proportional to:

Q_1 \propto \dfrac{\pi \cdot (1)^2}{1} = \pi
Q_2 \propto \dfrac{\pi \cdot (1)^2}{\frac{1}{2}} = 2\pi
Q_3 \propto \dfrac{\pi \cdot (2)^2}{2} = 2\pi
Q_4 \propto \dfrac{\pi \cdot (2)^2}{\frac{1}{2}} = 8\pi

Comparing these, Q_4 has the largest value, meaning the rod with r = 2 \, \text{cm}, \, l = \frac{1}{2} \, \text{m} will conduct the most heat.

Therefore, the correct answer is r = 2 \, \text{cm}, \, l = \frac{1}{2} \, \text{m}.

Answer 2