(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°C. 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°C 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\(^{-1}\)]
Explanation
(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
