2023 A/L Physics Past Paper · Question 36 (English Medium)

Answer: (5) Graph (5)

What the question asks

We want: the graph of the rate of heat flow $\dfrac{Q}{t}$ along the rod.

What each letter means:

$\dfrac{Q}{t}$ ($\text{W}$): the heat that flows through a cross-section each second.

$K$ ($\text{W m}^{-1}\,\text{K}^{-1}$): the thermal conductivity. It is the same in the three rods.

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

$\dfrac{\Delta\theta}{l}$ ($\text{K m}^{-1}$): the temperature gradient along the rod.

$r$ ($\text{m}$): the radius of a 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.

The first quote says that at steady state the rate of heat flow through any cross-section is the same, because the heat flows along the rod. The second quote says what the constant $K$ is.

For example, think of water flowing through a hose that gets thinner. If the hose does not leak, the same number of litres passes each point in a second. The water just moves faster in the thin part.

Heat in a lagged rod is the same. The same heat passes each second, and a thin rod needs a steeper temperature drop to carry it.

Required equations

$\dfrac{Q}{t} = K\,A\,\dfrac{\Delta\theta}{l}$

$A = \pi r^2$

$\dfrac{\Delta\theta}{l} = \dfrac{Q/t}{K\,A}$

The area goes as the square of the radius. The radii are in the ratio $4:2:1$, so the areas are in the ratio $16:4:1$.

Reasoning and calculation

Rod by rod:

The rate $\dfrac{Q}{t}$ is the same through each rod. The areas are $16:4:1$, so the gradients $\dfrac{\Delta\theta}{l}$ must be in the ratio $1:4:16$.

The thin rod has the steepest drop of temperature. The thick rod has the gentlest drop. The product of the area and the gradient does not change.

$1,\ 4,\ 16$

$\dfrac{\Delta\theta}{l} = \dfrac{16}{16},\ \dfrac{16}{4},\ \dfrac{16}{1}$

$16,\ 16,\ 16$

$\dfrac{Q}{t} = 1 \times 16 \times 1,\ 1 \times 4 \times 4,\ 1 \times 1 \times 16$

The shape of the graph:

The three pieces are at the same height, so they join into one straight horizontal line from $0$ to $3l$.

The line does not rise and does not fall. It does not touch zero, because heat does flow.

$k * a * g$

Reading the options

Now match it with the five graphs.

Graph (5) is a straight horizontal line. It matches.

Graph (1) and graph (2) rise. Graph (3) falls to zero. Graph (4) steps down. None of these match.

Why the other options fail

Graph (1) rises, but the rate is the same at each cross-section.

Graph (2) starts at zero, but heat enters the rod at $A$.

Graph (3) falls to zero, but no heat is lost from a lagged rod.

Graph (4) steps down, but the gradient changes to make the rate equal in each rod.

Conclusion

$\dfrac{Q}{t} = K\,A\,\dfrac{\Delta\theta}{l} = \text{constant}$

The rate of heat flow is the same along the whole composite rod. That is graph (5).

The answer is (5).

The quotes give the same rate through any cross-section at steady state, and what $K$ is. That the rate is $K A \dfrac{\Delta\theta}{l}$, with the gradient changing from rod to rod, is the rule of the lesson. The lagging means no heat leaves through the sides.

Teacher's note

At steady state in a lagged rod, the rate of heat flow is the same through each cross-section. A thinner rod does not carry less heat. It just has a steeper temperature drop.

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