2021 A/L Physics Past Paper · Question 30 (English Medium)

Answer: (2) $15\ \text{kJ}$

Given information

We know:

Coil: $R = 100\ \Omega$, $I = 1\ \text{A}$, $t = 5\ \text{min} = 300\ \text{s}$.

Air: constant pressure $p = 150\ \text{kPa} = 1.5 \times 10^5\ \text{Pa}$, volume from $0.4\ \text{m}^3$ to $0.5\ \text{m}^3$.

The container is well insulated, so no heat leaks out.

What each letter means:

$R$ ($\Omega$): resistance of the coil, how hard it is for current to pass.

$I$ ($\text{A}$): current, how much charge flows each second.

$t$ ($\text{s}$): time, how long the coil is on.

$p$ ($\text{Pa}$): pressure, how hard the air pushes on the piston.

$V$ ($\text{m}^3$): volume, the space the air fills.

$\Delta Q$ ($\text{J}$): heat that the coil gives to the air.

$\Delta W$ ($\text{J}$): work that the air does when it pushes the piston up.

$\Delta U$ ($\text{J}$): change in internal energy, the energy that stays inside the air.

What the question asks

We want: the change $\Delta U$ in the internal energy of the air.

The plan: find the heat from the coil, find the work done by the air, then use the first law.

Relevant theory

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

In a system, if the net exchange of heat is ∆Q, the change of internal energy is ∆U and the amount of work done is ∆W then,

∆ W = p.∆V (under constant pressure)

Required equations

$\Delta Q = I^2 R t$

$\Delta W = p\,\Delta V$

$\Delta Q = \Delta U + \Delta W$

Applying it to this question

For example:

Amma cooks rice in a pot with a heavy lid. The flame gives the pot heat. That is the heat from the coil, $\Delta Q$.

The steam pushes the heavy lid up. That is the work the air does when it pushes the piston up, $\Delta W$.

The rest of the heat makes the water hotter. That is the energy that stays inside the air, $\Delta U$.

$$\Delta Q = \Delta U + \Delta W$$

In words: $$heat\,from\,the\,flame = heat\,that\,warms\,the\,water + work\,that\,lifts\,the\,lid.\,Heat\,in = energy\,kept + work\,done$$

The coil gives more heat for a bigger current, a bigger resistance or a longer time. The air does more work for a bigger pressure or a bigger change of volume.

$30000$

$\Delta Q = 1^2 \times 100 \times 300$

$0.1$

$\Delta V = 0.5 - 0.4$

$15000$

$\Delta W = 1.5 \times 10^5 \times 0.1$

$\Delta U = \Delta Q - \Delta W = 30000 - 15000$

$\Delta Q$ and $\Delta W$ are both in joules, so $\Delta U$ is in joules: $15000\ \text{J} = 15\ \text{kJ}$.

Reasoning and calculation

Half of the heat goes into pushing the piston up. The other half stays in the air as internal energy.

$\Delta U = 15\ \text{kJ}$. The answer is (2).

Checks

$i * i * r * t - p * v$

Does it make sense?

The air expands, so it does work and $\Delta W$ is positive. So $\Delta U$ must be less than the heat supplied, $30\ \text{kJ}$.

The air is heated at constant pressure, so its temperature rises. So $\Delta U$ must be positive. $15\ \text{kJ}$ fits both.

Conclusion

$\Delta U = I^2Rt - p\Delta V = 15\ \text{kJ}$

$\Delta Q = 30\ \text{kJ}$, $\Delta W = 15\ \text{kJ}$.

$\Delta U = 30 - 15 = 15\ \text{kJ}$. The answer is (2).

In a heating question with a moving piston, the heat splits two ways: work on the piston and a rise in internal energy.

Change minutes to seconds before using $I^2Rt$.

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