2021 A/L Physics Past Paper · Question 36 (English Medium)

Answer: (5) $4$

What the question asks

We want: the rate of heat flow through $AB$ divided by the rate through $AC$, at the steady state.

We only need how the rate changes from rod to rod.

What each letter means:

$\dfrac{\Delta Q}{\Delta t}$ and $H$ ($\text{W}$): rate of heat flow, how much heat passes along the rod each second.

$K$ ($\text{W m}^{-1}\ \text{K}^{-1}$): thermal conductivity, how easily the material lets heat through.

$A$ ($\text{m}^2$): cross-sectional area of the rod.

$l$ ($\text{m}$): length of the rod.

$\theta_1$ and $\theta_2$ ($^\circ\text{C}$): temperatures at the two ends of the rod.

Relevant theory

In the steady state, the rate of flow of heat through any cross section like P or R should be the same because the flow of heat is axial.

The proportionality constant K here is known as the thermal conductivity of the material of the rod.

Required equations

$\dfrac{\Delta Q}{\Delta t} = K\dfrac{A(\theta_1 - \theta_2)}{l}$

Reasoning and calculation

What stays the same: every rod has the same area $A$, and the same length $l$ (the triangle is equilateral).

What changes: the conductivity $K$, and the temperature difference across the rod.

For example, put the end of a metal spoon in hot tea. Heat runs along the spoon to your hand.

A spoon of metal that conducts better passes more heat each second. That is a bigger $K$.

Hotter tea gives a bigger temperature difference between the two ends of the spoon. That also gives more heat each second.

In words: more heat flows each second when $K$ is bigger and when the temperature difference is bigger.

$$\dfrac{\Delta Q}{\Delta t} \propto K(\theta_1 - \theta_2)$$

Heat goes from $A$ to $B$ by two paths: straight along $AB$, or along $AC$ and then $CB$.

At the steady state the same rate goes through $AC$ and $CB$. They have the same $K$ and the same length.

So they share the $100\ ^\circ\text{C}$ equally, and $C$ is at $50\ ^\circ\text{C}$.

With $A$ and $l$ fixed:

Take the ratio of $AB$ to $AC$:

$$\dfrac{H_{AB}}{H_{AC}} = \dfrac{2K \times (100 - 0)}{K \times (100 - 50)}$$

Simplify:

$$\dfrac{H_{AB}}{H_{AC}} = \dfrac{200}{50} = 4$$

Reading the options

The value $4$ is option (5).

Option (4), $2$, uses only the conductivity. Options (1) and (2) are upside down.

Checks

Dimension check:

Both rates are $K\dfrac{A\,\Delta\theta}{l}$, in $\text{W}$. In the ratio, $A$, $l$ and the units cancel.

So the answer is a pure number, as the options are.

$(k2 * d1) / (k1 * d2)$

Does it make sense?

$AB$ wins twice: it conducts twice as well, and it has twice the temperature difference.

$2 \times 2 = 4$. So the ratio must be more than $2$.

Conclusion

$\dfrac{H_{AB}}{H_{AC}} = 4$

$\dfrac{H_{AB}}{H_{AC}} = \dfrac{2K \times 100}{K \times 50} = 4$.

The answer is (5).

Rods in series carry the same rate of heat flow. Equal rods in series share the temperature difference equally.

Find the temperature at the joint first, then use $K\dfrac{A\,\Delta\theta}{l}$.

Exam tip