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2024 WAEC Physics Theory (a)(i) What is a thermometric liquid? (ii) State the reason for the following design features...

Physics
WAEC 2024

(a)(i) What is a thermometric liquid?

(ii) State the reason for the following design features of a clinical thermometer. I. Narrow bore: II. Thin wall of the bulb 

(b) Distinguish between heat and temperature of an object in terms of the energy of a particle

(c) Explain why evaporation leads to cooling

(d) A kettle rated 2000W, contains water at 20ºC. The kettle is switched on and after two minutes, the water starts boiling. After another six minutes, 45% of the water in the kettle boils away. (i) Determine the specific latent heat of the vaporization of the water (ii) State one assumption made in your calculation 9d(i) above

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Explanation

(a) A thermometric liquid is a liquid used in thermometers to measure temperature. It expands and contracts uniformly with changes in temperature, allowing for accurate readings.

(ii) I. Narrow Bore
Reason: Allows precise measurement by controlling the movement of the thermometric liquid, ensuring quick and accurate readings

    II. Thin Wall of the Bulb
Reason: Enhances thermal conductivity, enabling faster heat transfer for quicker and more accurate temperature readings.

(b) Heat is a measure of the change in total internal energy in a body while temperature is a measure of the average kinetic energy of a molecule of the body.

(c) Evaporation leads to cooling for these reasons: 

          Energy Absorption: Molecules at the surface absorb energy to break free, often from the liquid and its surroundings.
          Loss of High-Energy Molecules: Higher-energy molecules evaporate, reducing the average kinetic energy of the remaining liquid.
          Temperature Decrease: As the average kinetic energy drops, the temperature of the liquid decreases, resulting in cooling.

(d)(i) Given: P = 2000W, \(\theta\) = 20ºC, t = 2 mins = 120 secs.

P x t = mc\(\Delta\)\(\theta\)

m = \(\frac{P \times t}{c \times \Delta \theta}\)

m = \(\frac{2000 \times 120}{4200 \times (100 - 20)}\) = 0.714 kg

Also, Pt = ml

l = \(\frac{Pt}{m}\)

where m = 45% of 0.714, t = 6mins = 360secs

l = \(\frac{2000 \times 60 \times 6}{0.714 \times 0.45}\) = 2.29 x 10\(^6\)Jkg\(^{-1}\)

(d)(i) Assumption made: there is no loss of heat to the surroundings, and the heat capacity of the material of the kettle is negligible.


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