This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Here are the solutions to the remaining parts of your question, Philip. 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. (ii) The effect of pressure on the boiling point of a liquid: Increasing pressure raises the boiling point of a liquid. c) (i) Step 1: Calculate the mass of hot and cold water delivered in 4 minutes. Mass of hot water (m_h) = Rate of hot water × Time m_h = 6 kg/min × 4 min = 24 kg Mass of cold water (m_c) = Rate of cold water × Time m_c = 15 kg/min × 4 min = 60 kg Step 2: 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 T_f be the final temperature. The specific heat capacity of water (c) is the same for both. m_h c (T_h - T_f) = m_c c (T_f - T_c) m_h (T_h - T_f) = m_c (T_f - T_c) Step 3: Substitute the values and solve for T_f. 24 kg (80 °C - T_f) = 60 kg (T_f - 20 °C) 1920 - 24 T_f = 60 T_f - 1200 1920 + 1200 = 60 T_f + 24 T_f 3120 = 84 T_f T_f = (3120)/(84) T_f ≈ 37.14 °C The final temperature of the mixture is 37.1 °C. (ii) I. Step 1: Determine the range of the faulty thermometer and the true Celsius scale. True Celsius scale: Lower Fixed Point (LFP_t) = 0 °C Upper Fixed Point (UFP_t) = 100 °C Range = 100 - 0 = 100 °C Faulty thermometer: Lower Fixed Point (LFP_f) = 0.5 °C Upper Fixed Point (UFP_f) = 90 °C Range = 90 - 0.5 = 89.5 °C Step 2: 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 - LFP_tUFP_t - LFP_t = Faulty Reading - LFP_fUFP_f - LFP_f Let R_t be the true temperature when the faulty thermometer reads R_f = 40 °C. (R_t - 0)/(100 - 0) = (40 - 0.5)/(90 - 0.5) (R_t)/(100) = (39.5)/(89.5) Step 3: Solve for the true temperature R_t. R_t = 100 × (39.5)/(89.5) R_t = 100 × 0.441340782 R_t ≈ 44.13 °C The temperature at normal pressure when the faulty thermometer reads 40 °C is 44.1 °C. 3 done, 2 left today. You're making progress.