A-LevelPhysicsD.C. circuits9702/52 · Oct/Nov 2019 · Q215 Marks
A student is investigating how the resistance of a thermistor varies with temperature. The thermistor is placed in water, as shown in Fig. 2.1. [Figure 2.1] The thermistor is connected to a battery with electromotive force (e.m.f.) E and negligible internal resistance. The current I in the thermistor is measured. The resistance R of the thermistor is then determined using the expression E R= I The experiment is repeated for different temperatures of the water. It is suggested that the resistance R of the thermistor and the thermodynamic temperature T are related by the equation R = pTq where p and q are constants.
- (a)A graph is plotted of lg R on the y-axis against lg T on the x-axis. Determine expressions for the gradient and the y-intercept. gradient = y-intercept =[1]
- (b)The value of E is 9.4 ± 0.1 V. Values of T, I and lg T are given in Fig. 2.2. T/K I/mA R/10³ Ω lg (T/K) lg (R/10³Ω) 303 1.0 ± 0.1 2.481 313 1.6 ± 0.1 2.496 323 2.4 ± 0.1 2.509 333 3.7 ± 0.1 2.522 343 5.5 ± 0.1 2.535 353 8.7 ± 0.1 2.548 Fig. 2.2 Calculate and record values of R/10³Ω and lg(R/10³Ω) in Fig. 2.2. Include the absolute uncertainties in R/10³Ω and lg (R/10³Ω).[4]
- (c)(i)Plot a graph of lg (R/10³Ω) against lg (T/K). Include error bars for lg (R/10³ Ω).[2]
- (c)(ii)Draw the straight line of best fit and a worst acceptable straight line on your graph. Both lines should be clearly labelled.[2]
- (c)(iii)Determine the gradient of the line of best fit. Include the absolute uncertainty in your answer. gradient =[2]
- (c)(iv)Determine the y-intercept of the line of best fit. Do not determine the absolute uncertainty. y-intercept =[1]
- (d)Using your answers to (a), (c)(iii) and (c)(iv), determine the values of p and q. You need not be concerned with units. Do not include the absolute uncertainties. p = q =[2]
- (e)Using your answers to (d), determine the thermodynamic temperature T when the resistance of the thermistor is 15 KΩ. T= K[1]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.
- [1] gradient = q and y-intercept = lg p
- [1] Values of R as above.
- [1] Values of Ig R as above.
- [1] Uncertainties in R from (±0.9 to ±1.2) to (±0.02 to ±0.03) and row 2 between ±0.40 and ±0.50 and row 4 between ±0.09 and ±0.10.
- [1] Uncertainties in Ig R consistent with uncertainties in R e.g. from ±0.05 to ±0.01.
- [1] Six points plotted correctly. Must be accurate to the nearest half a small square. Diameter of points must be less than half a small square.
- [1] Error bars in Ig R plotted correctly. All error bars must be plotted. Length of bar must be accurate to less than half a small square and symmetrical.
- [1] Line of best fit drawn. Upper end of line should pass between (2.500, 0.70) and (2.502, 0.70) and lower end of line should pass between (2.528, 0.30) and (2.532, 0.30). Do not accept line from first to last point.
- [1] Worst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). All error bars must be plotted.
- [1] Gradient determined with clear substitution of data points from the line of best fit into Δy / Δx. Distance between data points must be greater than half the length of the drawn line. Gradient must be negative.
- [1] uncertainty = gradient of line of best fit – gradient of worst acceptable line or uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)
- [1] y-intercept determined by substitution of correct point from the line of best fit into y = mx + c.
- [1] p determined from y-intercept. p (= 10y-intercept) = 10(c)(iv)
- [1] q = answer to (c)(iii) and given to 2 or 3 significant figures.
- [1] T determined from (d) or (c)(iii) and (c)(iv) with correct substitution shown. T = q√(R/p) T = q√(15/p) or lg T = (lg 15-lg p) / q lg T = (1.176-lg p) / q lg T = (lg 15-y-intercept) / gradient lg T = (1.176-(c)(iv)) / (c)(iii) T = 10^((1.176-(c)(iv))/(c)(iii))
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