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State one effect of pressure on the: (i) freezing point of a liquid; (ii) boiling point of a liquid.

Physics

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2 parts
c(i)

37.1 °C\text{37.1 °C}

c(ii)

44.1 °C\text{44.1 °C}

Step-by-step solution

4 steps

Here are the solutions to the remaining parts of your question, Philip.

Part b(i)

The effect of pressure on the freezing point of a liquid

For most substances, increasing pressure raises the freezing point. However, for water, increasing pressure lowers the freezing point.

Part b(ii)

The effect of pressure on the boiling point of a liquid

Increasing pressure raises the boiling point of a liquid.

Part c(i)

37.1 °C\text{37.1 °C}

Step 1: Calculate the mass of hot and cold water delivered in 4 minutes.

Mass of hot water (mhm_h) = Rate of hot water ×\times Time

mh=6 kg/min×4 min=24 kgm_h = 6 \text{ kg/min} \times 4 \text{ min} = 24 \text{ kg}

Mass of cold water (mcm_c) = Rate of cold water ×\times Time

mc=15 kg/min×4 min=60 kgm_c = 15 \text{ kg/min} \times 4 \text{ min} = 60 \text{ kg}
  1. Apply the principle of calorimetry (heat lost = heat gained).

    Assuming no heat loss to the surroundings, the heat lost by the hot water equals the heat gained by the cold water. Let TfT_f be the final temperature. The specific heat capacity of water (cc) is the same for both.

    mhc(Th−Tf)=mcc(Tf−Tc)m_h c (T_h - T_f) = m_c c (T_f - T_c) mh(Th−Tf)=mc(Tf−Tc)m_h (T_h - T_f) = m_c (T_f - T_c)
  2. Substitute the values and solve for TfT_f.

    24 kg(80 °C−Tf)=60 kg(Tf−20 °C)24 \text{ kg} (80 \text{ °C} - T_f) = 60 \text{ kg} (T_f - 20 \text{ °C}) 1920−24Tf=60Tf−12001920 - 24 T_f = 60 T_f - 1200 1920+1200=60Tf+24Tf1920 + 1200 = 60 T_f + 24 T_f 3120=84Tf3120 = 84 T_f Tf=312084T_f = \frac{3120}{84} Tf≈37.14 °CT_f \approx 37.14 \text{ °C}

    The final temperature of the mixture is 37.1 °C\boxed{\text{37.1 °C}}.

Part c(ii)

44.1 °C\text{44.1 °C}

I. Step 1: Determine the range of the faulty thermometer and the true Celsius scale.

True Celsius scale:
Lower Fixed Point (LFPtLFP_t) = 0 °C0 \text{ °C}
Upper Fixed Point (UFPtUFP_t) = 100 °C100 \text{ °C}
Range = 100−0=100 °C100 - 0 = 100 \text{ °C}

Faulty thermometer:
Lower Fixed Point (LFPfLFP_f) = 0.5 °C0.5 \text{ °C}
Upper Fixed Point (UFPfUFP_f) = 90 °C90 \text{ °C}
Range = 90−0.5=89.5 °C90 - 0.5 = 89.5 \text{ °C}

  1. Use the conversion formula for thermometer scales.

    The ratio of the temperature difference from the lower fixed point to the total range is constant for both scales:

    True Reading−LFPtUFPt−LFPt=Faulty Reading−LFPfUFPf−LFPf\frac{\text{True Reading} - LFP_t}{UFP_t - LFP_t} = \frac{\text{Faulty Reading} - LFP_f}{UFP_f - LFP_f}

    Let RtR_t be the true temperature when the faulty thermometer reads Rf=40 °CR_f = 40 \text{ °C}.

    Rt−0100−0=40−0.590−0.5\frac{R_t - 0}{100 - 0} = \frac{40 - 0.5}{90 - 0.5} Rt100=39.589.5\frac{R_t}{100} = \frac{39.5}{89.5}
  2. Solve for the true temperature RtR_t.

    Rt=100×39.589.5R_t = 100 \times \frac{39.5}{89.5} Rt=100×0.441340782R_t = 100 \times 0.441340782 Rt≈44.13 °CR_t \approx 44.13 \text{ °C}

    The temperature at normal pressure when the faulty thermometer reads 40 °C40 \text{ °C} is 44.1 °C\boxed{\text{44.1 °C}}.

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