2022 A/L Physics Past Paper · Question 40 (English Medium)

Answer: (2) Graph (2)

What the question asks

We want: the graph of the rate of heat flow $Q$ along the rods $AB$, $BC$ and $CE$.

What each letter means:

$Q$ ($\text{W}$): the rate of flow of heat along a rod.

$\theta_1$ ($\text{K}$): the temperature of end $A$. It is the higher one.

$\theta_2$ ($\text{K}$): the temperature of end $E$.

$k_1$ ($\text{W m}^{-1}\text{K}^{-1}$): the thermal conductivity of rods $AB$, $BC$ and $BD$.

$k_2$ ($\text{W m}^{-1}\text{K}^{-1}$): the thermal conductivity of rods $CD$, $CE$ and $DE$.

Relevant theory

What is happening:

Heat flows from the hot end $A$ to the cold end $E$ along the rods.

At the start, parts of the rods warm up and the flow changes. After a while the temperature at every point stops changing. That is the steady state.

At $B$ the heat meets two identical rods, $BC$ and $BD$. The two ways are the same, so the heat splits into two equal parts.

The network looks the same from $C$ as from $D$. So $C$ and $D$ are at the same temperature, and no heat passes along $CD$.

So the half that comes along $BC$ goes on along $CE$, and the half that comes along $BD$ goes on along $DE$.

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.

in the steady state, is proportional to, (1) The cross sectional area of the rod (A) (2) The temperature gradient along the rod

The rate of heat flow depends on the temperature gradient. A rod with the same temperature at both ends has no temperature gradient, so no heat flows along it. That is rod $CD$.

For example, water in a river that splits into two equal streams and joins again. Each stream carries half of the water. A small channel joining the two streams carries no water if the water is at the same height at both ends.

Heat in the rods is like that water. $BC$ and $BD$ are the two streams and $CD$ is the small channel.

Required equations

$Q_{AB} = Q$

$Q_{BC} = Q_{BD} = \dfrac{Q}{2}$

$Q_{CD} = 0$

$Q_{CE} = Q_{BC} = \dfrac{Q}{2}$

Reasoning and calculation

All the quantities here are heat flows in $\text{W}$. Two flows of $\dfrac{Q}{2}$ add up to $Q$ at $B$, so the total is the same on both sides of the junction.

To check, take sample numbers. Let $k_1 = 4$ and $k_2 = 2$ in units where a rod carries $k \times$ (temperature difference). Let $\theta_1 = 100$ and $\theta_2 = 0$.

Try the temperatures $\theta_B = 60$ and $\theta_C = \theta_D = 40$. Then see if the heat going into each junction equals the heat going out.

$160$

$Q_{AB} = k_1(\theta_1 - \theta_B) = 4 \times (100 - 60)$

$80$

$Q_{BC} = k_1(\theta_B - \theta_C) = 4 \times (60 - 40)$

$0$

$Q_{CD} = k_2(\theta_C - \theta_D) = 2 \times (40 - 40)$

$Q_{CE} = k_2(\theta_C - \theta_2) = 2 \times (40 - 0)$

At $B$: $160$ comes in and $80 + 80 = 160$ goes out along $BC$ and $BD$. At $C$: $80$ comes in along $BC$ and $80$ goes out along $CE$, with $0$ along $CD$. Heat in equals heat out at $B$ and at $C$, and $D$ is the same as $C$. So these temperatures are the steady state.

So $Q_{AB} = 160$, and $Q_{BC} = Q_{CE} = 80$. The graph is level along $AB$, drops at $B$ to half, and is level again along $BC$ and $CE$ at the same value.

Check the five graphs.

Graph (1) has no drop. Graph (3) drops again at $C$. Graphs (4) and (5) slope or curve.

Graph (2) drops at $B$ and then stays level to $E$. It matches.

Checks

$k2 * (tc - te)$

Does it make sense?

The heat that leaves $A$ must reach $E$, because the rods are lagged and nothing is lost on the way. At $E$ the two rods $CE$ and $DE$ bring $80 + 80 = 160$. That is the same as the $160$ that left $A$.

The values of $k_1$ and $k_2$ do change the temperatures at $B$, $C$ and $D$. They do not change the shape of the graph.

Conclusion

$\text{graph }(2)$

The heat flow drops to half at $B$ and then stays level along $BC$ and $CE$. The graph is (2).

The answer is (2).

At a junction with two identical rods the heat splits equally.

A rod with the same temperature at both ends carries no heat. Use that to remove it from the picture.

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