(a) (i) Explain latent heat.
(ii) State two factors that affect the rate of evaporation of a liquid
(b) Explain each of the following observations:
(i) On a dry day, water in a clay pot is cooler than water in a closed plastic container;
(ii) Food gets cooked faster in a pressure cooker than in an ordinary cooking pot.
(c) State two effects of heat on a substance.
(d) A 40 V electric heater is used to supply a current of 12 A for 1400 s to a body of mass 1.5 kg at the melting point of the body. The body melts and its temperature rises through 60\(^o\)C in an extra 72 s. Determine the:
(i) latent heat of fusion of the body;
(ii)specific heat capacity of the body.
(a)(i) Explanation of latent heat
When a substance is changing state, it absorbs/gives out thermal/heat energy which does not result in temperature change. This heat absorbed or given out during change of state is known as latent heat.
(ii) Factors that affect the rate of evaporation of a liquid
- exposed surface area of the liquid
- temperature of the liquid
- wind / draught
- humidity / amount of water vapour
- nature of the liquid
- air pressure.
(b) Explanation of observations
(i) Water in a clay pot is cooler than water in a closed plastic container
The clay pot has pores which allow evaporation to take place. In the process of evaporation, molecules take latent heat from the bulk of the water, leaving the water cooler; but the plastic container does not have pores (to allow for evaporation).
(ii) Food gets cooked faster in a pressure cooker than in an ordinary cooking pot
With the pressure cooker, increased pressure raises the boiling point of water within a short period. This leads to faster cooking. Ordinary cooking pot operates at normal atmospheric pressure.
(c) Effects of heat on a substance
Change of/in:
- state
- temperature
- dimension
- colour
- electrical resistance
(d) (i) Determination of latent heat of fusion of the body
Heat supplied by electric heater in the first 1400 s
IVt\(_1\) = mlf
mlf = 12 x 40 x 1400
= 6.72 x 105J
(ii) Determination of Specific heat capacity of the body
IVt\(_2\) = mc\(\Delta \theta\)
c = \(\frac{IVt_2}{m \Delta \theta}\) = \(\frac{12 \times 40 \times 72}{1.5 \times 60}\)
= 384 Jk\(^{-1}\) K\(^{-1}\)
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