| Date | May 2017 | Marks available | 1 | Reference code | 17M.2.SL.TZ1.4 |
| Level | Standard level | Paper | Paper 2 | Time zone | Time zone 1 |
| Command term | Calculate | Question number | 4 | Adapted from | N/A |
Question
A heater in an electric shower has a power of 8.5 kW when connected to a 240 V electrical supply. It is connected to the electrical supply by a copper cable.
The following data are available:
Length of cable = 10 m
Cross-sectional area of cable = 6.0 mm2
Resistivity of copper = 1.7 × 10–8 Ω m
Calculate the current in the copper cable.
Calculate the resistance of the cable.
Explain, in terms of electrons, what happens to the resistance of the cable as the temperature of the cable increases.
The heater changes the temperature of the water by 35 K. The specific heat capacity of water is 4200 J kg–1 K–1.
Determine the rate at which water flows through the shower. State an appropriate unit for your answer.
Markscheme
I «=\(\frac{{8.5 \times {{10}^3}}}{{240}}\)» =35«A»
R = \(\frac{{1.7 \times {{10}^{ - 8}} \times 10}}{{6.0 \times {{10}^{ - 6}}}}\)
= 0.028 «Ω»
Allow missed powers of 10 for MP1.
«as temperature increases» there is greater vibration of the metal atoms/lattice/lattice ions
OR
increased collisions of electrons
drift velocity decreases «so current decreases»
«as V constant so» R increases
Award [0] for suggestions that the speed of electrons increases so resistance decreases.
recognition that power = flow rate × cΔT
flow rate «\( = \frac{{{\text{power}}}}{{c\Delta T}}\)» \( = \frac{{8.5 \times {{10}^3}}}{{4200 \times 35}}\)
= 0.058 «kg s–1»
kg s−1 / g s−1 / l s−1 / ml s−1 / m3 s−1
Allow MP4 if a bald flow rate unit is stated. Do not allow imperial units.
