(a) List two factors each that affect heat loss by:
(i) radiation;
(ii) convection.
(b) State two factors that determine the quantity of heat in a body.
(c) Explain the statement: The vecilic latent heat of vaporization of mercury is 2.72 x 10\(^5\) Jkg\(^{-1}\).
(d)A jug of heat capacity 250 Jkg\(^{-1}\) contains water at 28^\circC. An electric heater of resistance 35\(\Omega\) connected to a 220 V source is used to raise the temperature of the water until it boils at 100^\circC in 4 minutes. After. another 5 minutes, 300 g of water has evaporated. Assuming no heat is lost to the surroundings, calculate the:
(i) mass of water in the jug before heating;
(ii) specific latent heat of vaporization of steam. [Specific heat capacity of water = 4200 kg\(^{-1}\)K\(^{-1}\)]
Verified Correct Answer: .
Physical Principles & Numerical Solution
(a) (i) Radiation:
Surface area, Temperature.
(ii) Convection:
Nature, Density/Viscosity of the fluid, Thermal conductivity of fluid, Specific heat capacity of fluid and Exposed surface area.
(b) Factors that determine the quantity of heat in a body: Heat/thermal capacity and. Temperature
(c) It means that 2.72 x 105J of heat energy is required/ needed to change I kg of mercury at its boiling point to vapour (without temperature change)
(d) (i) heat supplied by heater = ivt = \(\frac{v^2}{R}\)t
heat gained by water = M\(_w\)C\(_w\)(\(\theta_2\) - \(\theta_1\))
where M\(_w\) = mass of water
C\(_w\) = s.h.c of water
\(\theta_2\) = final temp. of water
\(\theta_1\) = initial temp. of water
Heat gained by jug = C\(_j\) (\(\theta_2\) - \(\theta_1\))
\(\frac{v^2t}{R}\) = M\(_w\)C\(_w\) (\(\theta_2\) - \(\theta_1\))
= \(\frac{220^2 \times 4 \times 60}{35}\)
= M\(_w\) x 4200[100 - 28] + 250[100 - 28]
M\(_w\) = 1.038kg
Source: Myschool Past Questions Archive (Record #52871)