A student is investigating the emf and internal resistance of a cell, using the circuit shown. The ammeter and voltmeter are ideal.
The graph shows the variation of the voltmeter reading \(V\) with the ammeter reading \(I\).
(a) Explain why \(V\) changes when the resistance of the variable resistor is changed.
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(b) Show that the internal resistance of the cell is about \(0.8\,\Omega\).
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(c) Determine the emf of the cell.
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(d) The student then replaces the variable resistor with an electric heating element in order to investigate the thermal output of the system. The heating element is used to raise the temperature of the water to the boiling point.
After the boiling point is reached, \(12\,\mathrm{g}\) of the water are vaporized. The specific latent heat of vaporization of water is \(2.3\times10^{6}\,\mathrm{J\,kg^{-1}}\).
(i) Calculate the change in entropy of the vaporized water.
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(ii) Explain why this entropy change is positive.
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Most-appropriate topic codes (IB Physics 2025):
(a) Correct Answer:
Changing the resistance of the variable resistor changes the current \(I\) in the circuit.
For a cell with emf \(\varepsilon\) and internal resistance \(r\), the terminal potential difference is
\(V=\varepsilon-Ir\)
Therefore, as \(I\) changes, the terminal voltage \(V\) also changes.
(b) Correct Answer: \( \boxed{0.75\,\Omega\approx0.8\,\Omega} \)
The equation for the terminal potential difference is
\(V=\varepsilon-Ir\)
Comparing this with \(y=mx+c\), the gradient of the \(V\)-\(I\) graph is \(-r\).
Using two points from the graph, approximately \((3\,\mathrm{A},22.5\,\mathrm{V})\) and \((10\,\mathrm{A},17.2\,\mathrm{V})\),
\(r=\dfrac{22.5-17.2}{10-3}\)
\(r\approx0.75\,\Omega\)
Therefore,
\(\boxed{r\approx0.8\,\Omega}\).
(c) Correct Answer: \( \boxed{24.7\,\mathrm{V}} \)
From
\(V=\varepsilon-Ir\)
when \(I=0\), the equation becomes
\(V=\varepsilon\)
Therefore, the emf is the \(V\)-axis intercept of the graph.
Extrapolating the line to \(I=0\) gives approximately
\(\boxed{\varepsilon\approx24.7\,\mathrm{V}}\)
(d)(i) Correct Answer: \( \boxed{74\,\mathrm{J\,K^{-1}}} \)
The heat energy supplied to vaporize the water is
\(Q=mL\)
where \(m=12\,\mathrm{g}=0.012\,\mathrm{kg}\).
Therefore,
\(Q=(0.012)(2.3\times10^{6})\)
\(Q=2.76\times10^{4}\,\mathrm{J}\)
For a phase change at constant temperature,
\(\Delta S=\dfrac{\Delta Q}{T}\)
The boiling point of water is \(T=373\,\mathrm{K}\), so
\(\Delta S=\dfrac{2.76\times10^{4}}{373}\)
\(\Delta S\approx74\,\mathrm{J\,K^{-1}}\)
Hence, \( \boxed{\Delta S\approx74\,\mathrm{J\,K^{-1}}} \).
(d)(ii) Correct Answer:
When liquid water changes into steam, the particles become much more widely separated and can occupy a greater number of possible arrangements, or microstates.
This corresponds to greater disorder and a greater number of ways to distribute the energy among the particles.
Therefore, the entropy increases and \( \boxed{\Delta S>0} \).
