(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 th...

Assessment: WAEC SSCE - Physics - 2024 (Essay) Subject: Physics

Question 1 Report

(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

Answer Details

(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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