2023 A/L Physics Past Paper · Question 18 (English Medium)

Answer: (4) $2\ \text{min}$

What the question asks

What we know:

The first heater has a wire of length $L$ and diameter $d$. It takes $4\ \text{min}$ to warm the water by $40\ ^\circ\text{C}$.

The second heater has a wire of the same material, with length $2L$ and diameter $2d$. It warms the same amount of water by the same $40\ ^\circ\text{C}$.

No heat is lost to the surroundings.

The question does not give the supply voltage. We take it as the same for both heaters, as if both were plugged into one socket.

We want: the time $t_2$ the second heater takes.

What each letter means:

$L$ ($\text{m}$): the length of the first wire.

$d$ ($\text{m}$): the diameter of the first wire.

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

$\rho$ ($\Omega\ \text{m}$): the resistivity of the material.

$R$ ($\Omega$): the resistance of a wire.

$V$ ($\text{V}$): the potential difference across the heater.

$W$ ($\text{J}$): the heat energy that the water needs.

$t$ ($\text{min}$): the time the heater takes.

Relevant theory

(1) When the length of a conductor increases its resistance increases. (2) When the area of cross-section of a conductor increases its resistance decreases.

The constant p represents the contribution from the material of the conductor to its resistance and is known as the "resistivity" of the material.

Heat is mainly liberated when a current passes through a resistor and this is known as "joule heating".

Passive resistors are those resistors in which the electrical energy flowing through them gets completely converted into heat.

The first two quotes tell us how the resistance of a wire depends on its length, its area and its material. The last two quotes say that the electrical energy that flows through a heater wire is completely changed into heat.

So the heat the water gets is the electrical energy $W = VIt$ that the wire uses.

We also know the resistance of the wire, $V = IR$. Put $I = \dfrac{V}{R}$ in $W = VIt$, and we get $W = \dfrac{V^2t}{R}$.

The same water must get the same heat, and $V$ is the same. So the time $t$ goes with the resistance $R$.

Required equations

$R = \rho\dfrac{l}{A}$

$A = \dfrac{\pi d^2}{4}$

$W = VIt$

$I = \dfrac{V}{R}$

$W = \dfrac{V^2 t}{R}$

Reasoning and calculation

What stays the same: the material, so $\rho$ is the same. The water and the $40\ ^\circ\text{C}$ rise are the same, so the heat $W$ is the same. The voltage $V$ is the same.

What changes: the length and the diameter of the wire. So $R$ changes, and with it the time $t$.

For example, think of two taps filling the same bucket from one water supply.

A long, narrow pipe lets the water through slowly, so the bucket takes a long time. A short, wide pipe lets a lot through, so the bucket fills fast.

A heater wire is like the pipe. A wire that is harder to push the current through lets less current flow, so it needs more time to give the same heat.

Write the time for each heater, then divide.

$$t_1 = \dfrac{WR_1}{V^2} \qquad t_2 = \dfrac{WR_2}{V^2}$$

$$\dfrac{t_2}{t_1} = \dfrac{R_2}{R_1}$$

The $W$ and the $V$ cancel.

The resistance of each wire is $R = \rho\dfrac{l}{A}$ with $A = \dfrac{\pi d^2}{4}$. The $\rho$ and the $\pi$ cancel too, so

$$\dfrac{R_2}{R_1} = \dfrac{l_2}{l_1} \times \left(\dfrac{d_1}{d_2}\right)^2$$

The second wire is twice as long, and twice as wide.

$$\dfrac{l_2}{l_1} = 2 \qquad \dfrac{d_1}{d_2} = \dfrac{1}{2}$$

Put them in:

$$\dfrac{R_2}{R_1} = 2 \times \left(\dfrac{1}{2}\right)^2 = 2 \times \dfrac{1}{4} = \dfrac{1}{2}$$

So the second time is half of the first:

$$t_2 = \dfrac{1}{2} \times 4\ \text{min} = 2\ \text{min}$$

Reading the options

The second heater takes $2\ \text{min}$. That is option (4).

Option (2) is what the wider wire gives without the longer length, and option (5) is what the longer wire gives without the wider wire.

Checks

Dimension check:

$\dfrac{t_2}{t_1}$ is a time divided by a time, and $\dfrac{R_2}{R_1}$ is a resistance divided by a resistance. Both are plain numbers, so they can be put equal.

$t1 * l / (d * d)$

Does it make sense?

The second wire is half the resistance, so at the same voltage twice as much current flows. The heater gives heat twice as fast.

The same heat at twice the rate takes half the time. That is $2\ \text{min}$.

Conclusion

$t_2 = 2\ \text{min}$

$\dfrac{t_2}{t_1} = \dfrac{R_2}{R_1} = 2 \times \dfrac{1}{4} = \dfrac{1}{2}$, so $t_2 = 2\ \text{min}$.

The answer is (4).

For the same heat and the same voltage, the time goes with the resistance. So find $\dfrac{R_2}{R_1}$.

Length goes on top, and the diameter goes underneath as a square: $\dfrac{R_2}{R_1} = \dfrac{l_2}{l_1}\left(\dfrac{d_1}{d_2}\right)^2$.

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