IB · B · data analysis
The particulate nature of matter · Question 7
The particulate nature of matter · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- demanding
- Marks
- 11
- Topics
- 1
- Answer
- Complete
A student determines the specific heat capacity of a liquid. She places 0.500 kg of the liquid in a well-insulated container with an electric heater of constant power 50.0 W and records the temperature at intervals.
| t / s | 0 | 120 | 240 | 360 | 480 |
|---|---|---|---|---|---|
| θ / °C | 20.0 | 22.9 | 25.7 | 28.6 | 31.4 |
- (a)
Determine Determine the gradient of a graph of temperature against time, and explain why the gradient rather than a single pair of readings should be used.
3 marks - (b)
Determine Determine the specific heat capacity of the liquid.
3 marks - (c)
Outline Outline why the container being well insulated matters more at the end of the experiment than at the beginning.
2 marks - (d)
Suggest The heater and the container itself also absorb energy. Suggest how this affects the value obtained for c, and suggest one way of accounting for it.
3 marks
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(a)
Determine Determine the gradient of a graph of temperature against time, and explain why the gradient rather than a single pair of readings should be used.
Read the data before explaining it. Quote the relevant values or trend, show the comparison or calculation, and then connect that numerical evidence to the physical conclusion—including uncertainty or anomalies when they matter.
- 1
gradient = (31.4 − 20.0)/480 = 0.0238 °C s⁻¹
- 2
the gradient uses all five readings, so random errors in individual temperature measurements partly cancel
- 3
and a straight line confirms that the rate of heating is constant, which the calculation assumes
(b)
Determine Determine the specific heat capacity of the liquid.
Read the data before explaining it. Quote the relevant values or trend, show the comparison or calculation, and then connect that numerical evidence to the physical conclusion—including uncertainty or anomalies when they matter.
- 1
Constant heater power satisfies P = mc(dθ/dt), where dθ/dt is the measured temperature gradient.
- 2
c = P / (m × gradient) = 50.0 / (0.500 × 0.0238)
- 3
c = 4.2 × 10³ J kg⁻¹ K⁻¹
(c)
Outline Outline why the container being well insulated matters more at the end of the experiment than at the beginning.
Read the data before explaining it. Quote the relevant values or trend, show the comparison or calculation, and then connect that numerical evidence to the physical conclusion—including uncertainty or anomalies when they matter.
- 1
the rate of energy transfer to the surroundings depends on the temperature difference between the liquid and the room
- 2
at the start the liquid is near room temperature and loses almost nothing; by the end it is 11 °C above the room and the losses are largest
(d)
Suggest The heater and the container itself also absorb energy. Suggest how this affects the value obtained for c, and suggest one way of accounting for it.
Read the data before explaining it. Quote the relevant values or trend, show the comparison or calculation, and then connect that numerical evidence to the physical conclusion—including uncertainty or anomalies when they matter.
- 1
some of the 50.0 W raises the temperature of the heater and container rather than the liquid
- 2
the liquid therefore warms more slowly than the model assumes, the gradient is smaller, and the calculated value of c is too large
- 3
account for it by repeating the experiment with a different mass of liquid and using the difference, or by determining the thermal capacity of the container separately and subtracting it
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