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A mass of 5 kg of water is heated with a 500 W heater for 20 minutes. (i) Calculate the heat supplied. (ii) Determine the change in temperature. (iii) Explain why water is used as a cooling liquid.

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7. (e) A mass of 5 kg5 \text{ kg} of water is heated with a 500 W500 \text{ W} heater for 20 minutes20 \text{ minutes}.

(i) Calculate the heat supplied by the heater supplied during this time.

  1. Convert time from minutes to seconds.

    t=20 minutes×60 s/minute=1200 st = 20 \text{ minutes} \times 60 \text{ s/minute} = 1200 \text{ s}
  2. Use the formula for heat supplied by a heater, Q=P×tQ = P \times t.

    Q=500 W×1200 sQ = 500 \text{ W} \times 1200 \text{ s} Q=600000 JQ = 600000 \text{ J}

    The heat supplied by the heater is 600000 J\boxed{600000 \text{ J}}.

    (ii) Given that the specific heat capacity of water is 4200 J kg−1K−14200 \text{ J kg}^{-1} \text{K}^{-1}, determine the change in temperature of the water during this time.

  3. Use the formula for heat absorbed, Q=mcΔTQ = mc\Delta T.

    We have Q=600000 JQ = 600000 \text{ J} (from part i), m=5 kgm = 5 \text{ kg}, and c=4200 J kg−1K−1c = 4200 \text{ J kg}^{-1} \text{K}^{-1}.

  4. Rearrange the formula to solve for ΔT\Delta T.

    ΔT=Qmc\Delta T = \frac{Q}{mc} ΔT=600000 J5 kg×4200 J kg−1K−1\Delta T = \frac{600000 \text{ J}}{5 \text{ kg} \times 4200 \text{ J kg}^{-1} \text{K}^{-1}} ΔT=60000021000 K\Delta T = \frac{600000}{21000} \text{ K} ΔT≈28.57 K\Delta T \approx 28.57 \text{ K}

    The change in temperature of the water is approximately 28.57 K\boxed{28.57 \text{ K}}.

    (iii) Explain why water is often used as a cooling liquid in car radiators.
    Water is often used as a cooling liquid in car radiators because it has a high specific heat capacity. This means it can absorb a large amount of heat energy from the engine without a significant rise in its own temperature, effectively cooling the engine.

    7. (f) A student used an un-calibrated thermometer in the laboratory to measure the temperature of a given liquid. He first put it in pure melting ice and the mercury thread was 10 cm10 \text{ cm} long. He then removed it and put it in steam from boiling water, the mercury thread was 50 cm50 \text{ cm} long. Then he puts it in the liquid and the mercury thread was 35 cm35 \text{ cm} long.

    (i) Define temperature.
    Temperature is a measure of the average kinetic energy of the particles (atoms or molecules) within a substance, indicating its degree of hotness or coldness.

    (ii) State two properties that make mercury suitable for use in thermometers.

    1. Mercury has a uniform and visible expansion over a wide range of temperatures.
    2. It has a high boiling point and a low freezing point, allowing it to measure a broad range of temperatures in liquid form.
    3. It is a good conductor of heat, allowing it to quickly respond to temperature changes. (Any two of these are suitable).

    (iii) Calculate the temperature of the liquid.

  5. Identify the given lengths and corresponding temperatures.

    Length at 0∘C0^\circ C (melting ice), L0=10 cmL_0 = 10 \text{ cm}.
    Length at 100∘C100^\circ C (boiling water), L100=50 cmL_{100} = 50 \text{ cm}.
    Length in the liquid, Lx=35 cmL_x = 35 \text{ cm}.

  6. Use the formula for calculating temperature with an un-calibrated thermometer.

    Tx=Lx−L0L100−L0×100∘CT_x = \frac{L_x - L_0}{L_{100} - L_0} \times 100^\circ C
  7. Substitute the values and calculate TxT_x.

    Tx=35 cm−10 cm50 cm−10 cm×100∘CT_x = \frac{35 \text{ cm} - 10 \text{ cm}}{50 \text{ cm} - 10 \text{ cm}} \times 100^\circ C Tx=25 cm40 cm×100∘CT_x = \frac{25 \text{ cm}}{40 \text{ cm}} \times 100^\circ C Tx=58×100∘CT_x = \frac{5}{8} \times 100^\circ C Tx=0.625×100∘CT_x = 0.625 \times 100^\circ C Tx=62.5∘CT_x = 62.5^\circ C

    The temperature of the liquid is 62.5∘C\boxed{62.5^\circ C}.

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