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    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:

    Question
    SLPaper 2

    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

    1.

    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.

    [4]
    Verified
    Solution

    recognition that power === flow rate ×cΔT\times c\Delta T×cΔT 1 mark

    flow rate «=powercΔT»=8.5×1034200×35«=\frac{\text{power}}{c\Delta T} »=\frac{8.5 \times 10^{3}}{4200 \times 35} «=cΔTpower​»=4200×358.5×103​ 1 mark

    =0.058≪kgs−1»=0.058 \ll \mathrm{kgs}^{-1} » =0.058≪kgs−1» 1 mark

    kg s−1/g s−1/l s−1/ml s−1/m3s−1\mathrm{kg\ s}^{-1} / \mathrm{g\ s}^{-1} / \mathrm{l\ s}^{-1} / \mathrm{ml\ s}^{-1} / \mathrm{m}^{3} \mathrm{s}^{-1}kg s−1/g s−1/l s−1/ml s−1/m3s−1 1 mark

    2.

    Calculate the resistance of the cable.

    [2]
    Verified
    Solution

    R=1.7×10−8×106.0×10−6R=\frac{1.7 \times 10^{-8} \times 10}{6.0 \times 10^{-6}} R=6.0×10−61.7×10−8×10​ 1 mark

    =0.028«Ω»=0.028 «\Omega » =0.028«Ω» 1 mark

    3.

    Explain, in terms of electrons, what happens to the resistance of the cable as the temperature of the cable increases.

    [3]
    Verified
    Solution

    «as temperature increases» there is greater vibration of the metal atoms/lattice/lattice ions

    **OR **

    increased collisions of electrons 1 mark

    drift velocity decreases «so current decreases» 1 mark

    «as V constant so» RRR increases 1 mark

    Award [0] for suggestions that the speed of electrons increases so resistance decreases.

    4.

    Calculate the current in the copper cable.

    [1]
    Verified
    Solution

    I«=8.5×103240»=35I «=\frac{8.5 \times 10^{3}}{240} »=35I«=2408.5×103​»=35 A 1 mark

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