Draw the best-fit line for the data points.

Calculate the gradient of the graph when $T = 291{\text{ K}}$.

State the unit for your answer to (b)(i).

The uncertainty in the resistance value is 5%. The uncertainty in the temperature is negligible. On the graph, draw error bars for the data point at $T = 283{\text{ K}}$ and for the data point at $T = 319{\text{ K}}$.

Calculate the power dissipated by the thermistor at $T = 283{\text{ K}}$.

Determine the uncertainty in the power dissipated by the thermistor at $T = 283{\text{ K}}$.

smooth curve that passes within $\pm 0.5$ squares of all data points;

a tangent drawn at $[291,{\text{ }}5.2]$ and selection of two extreme points on the tangent that use $\Delta R > 3.5{\text{ }}\Omega$; } (judge by eye)

gradient magnitude determined as $0.20 \pm 0.02$;

negative value given;

$\Omega {\text{ }}{{\text{K}}^{ - 1}}$;

correct error bar for 283 K (total length of bar 3–5 squares, centred on point);

correct error bar for 319 K (total length of bar 0.5–2 squares, centred on point);

substituting ${I^2}R = \left( {{{[0.78 \times {{10}^{ - 3}}]}^2} \times 7.5} \right) = 4.5 \times {10^{ - 6}}{\text{ W}}$$\,\,\,$or$\,\,\,$$4.6 \times {10^{ - 6}}{\text{ W}}$;

fractional uncertainty in ${I^2} = 2 \times \frac{{0.01}}{{0.78}}$ (= 0.026$\,\,\,$or$\,\,\,$2.6%);

uncertainty in power $\left( { = [0.026 + 0.05] \times 4.6 \times {{10}^{ - 6}}} \right) = 0.34 \times {10^{ - 6}}{\text{ W}}$ to $0.35 \times {10^{ - 6}}{\text{ W}}$;

answer rounded to 1 significant figure;

or

uncertainty in ${I^2} = 2 \times 1.3\% /0.026$;

total uncertainty in ${\text{P}} = 7.6\% /0.076$;

answer rounded to 1 significant figure;

Very poorly executed. Few SL candidates knew how to draw an acceptable tangent.