Prelim H2P4 - soln

1 1 In this experiment, you will investigate how the resistance R of a thermistor varies with its temperature θ. (a) Construct the circuit shown in Fig. 1.1. Fig. 1.1 (b) Immerse the thermistor in a beaker of water. Measure and record the current I, potential difference V and temperature θ. I = .............................. V = .............................. θ = ……....................[1] (c) (i) Arrange the apparatus so that the water may be heated. Use the stand and clamp to ensure that the wires connected to the thermistor are kept well away from the source of heating. (ii) Use the heating apparatus to raise the temperature of the water by about 10 0C. After the temperature has stabilized, measure and record the new values of I, V and θ. I = .............................. V = .............................. θ = ……....................... 9749/04/ASRJC/2021Prelim [Turn Over M1 For Examiner’s Use V mA 11.1 mA 2.08 V 30.0 0C 17.4 mA 2.08 V 43.0 0C

2 (iii) Repeat (c)(ii) for θ varying from room temperature to 80 °C. Tabulate these results. Include the results from (b), (c)(ii), all the values of R, where R = V / I, and all the values of thermodynamic temperature T, where T = θ + 273. [6] θ / °C V / V I / mA R / Ω T / K ln (R/ Ω) 1 T / 10 3‒ K‒ 1 30.0 2.08 11.1 187 303.0 5.23 3.300 43.0 2.08 17.4 120 316.0 4.79 3.165 50.0 2.08 23.5 88.5 323.0 4.48 3.096 60.0 2.08 31.6 65.8 333.0 4.19 3.003 70.0 2.07 41.9 49.4 343.0 3.90 2.915 80.0 2.07 54.3 38.1 353.0 3.64 2.833 (d) The formula which relates R and T is where A is a constant, E is an energy characteristic of the thermistor, and k = 1.38 × 10 23‒ J K 1‒ . Plot a suitable graph to determine values for E and A. [7] R= Ae E kT ln R= E k ( 1 T )+ ln A Plot a graph of ln R against 1/T where gradient = E/k, and y-intercept = ln A. Hence Substituting (3.440 × 10 3‒ , 5.72) into y = mx + c: y-intercept c = 5.72 – 3.46 × 103 (3.440 × 10 3‒ ) = – 6.18 ln A = 6.18‒ A = e 6.18‒ = 2.07 × 10 3‒ Ω E = …………..……….. 9749/04/ASRJC/2021Prelim For Examiner’s Use P2 P3 P1 M2 A2 A4 A5 A3 A1 Gradient = 4.77 × 10 20‒ J 2.07 × 10 3‒ Ω

3 A = ……..…………….. 9749/04/ASRJC/2021Prelim [Turn Over For Examiner’s Use P4 P5 P6 3.60 3.80 4.00 5.80 4.20 5.60 5.40 4.40 5.20 4.60 4.80 5.00 2.800 2.900 3.000 3.100 3.200 3.300 3.400 3.500 (2.880, 3.78) (3.440, 5.72) x x x x x x ln (R/)

4 (e) Comment on any anomalous data or results you may have obtained. Explain your answer. ……..…………………………………………………………………………………….………… ……………..…………………………………………………………………………….………… ……………………...………………………………………..…………………………….……[1] [Total: 15 marks] 9749/04/ASRJC/2021Prelim M3 There are no anomalous data as all th