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.