(a) (i) Define relative density.
(ii) List three characteristics of pressure in a liquid.
(b) The horizontal door of a submarine at a depth of 500m has an area of 0.4m\(^{2}\). Calculate the force exerted by the sea water on the door at this depth. [Relative density of sea water = 1.03 ] [Atmospheric pressure =- 1.01 x 10\(^{5}\) Nm\(^{-2}\)] [Density of pure water = 1000kgm\(^{-3}\)] [g = 10ms\(^{-2}\)]
(c) (i) List three effects of heat other than expansion.
(ii) Explain saturated vapour pressure.
(iii) A heating coil of resistance 20 \(\Omega\) connected to a 220 V source is used to boil a certain quantity of water in a container of heat capacity 100 J kg\(^{-1}\) for 2 minutes. If the initial temperature of the water is 40° C, calculate the mass of the water in the container. [specific heat capacity of water = 4.2 x 10\(^3\) Jkg\(^{-1}\) K\(^{-1}\)] [assume boiling point of water = 100°C]
(ii) List three characteristics of pressure in a liquid.
(b) The horizontal door of a submarine at a depth of 500m has an area of 0.4m\(^{2}\). Calculate the force exerted by the sea water on the door at this depth. [Relative density of sea water = 1.03 ] [Atmospheric pressure =- 1.01 x 10\(^{5}\) Nm\(^{-2}\)] [Density of pure water = 1000kgm\(^{-3}\)] [g = 10ms\(^{-2}\)]
(c) (i) List three effects of heat other than expansion.
(ii) Explain saturated vapour pressure.
(iii) A heating coil of resistance 20 \(\Omega\) connected to a 220 V source is used to boil a certain quantity of water in a container of heat capacity 100 J kg\(^{-1}\) for 2 minutes. If the initial temperature of the water is 40° C, calculate the mass of the water in the container. [specific heat capacity of water = 4.2 x 10\(^3\) Jkg\(^{-1}\) K\(^{-1}\)] [assume boiling point of water = 100°C]
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Correct Answer: Option n
Explanation:
(a)(i) Relative density is the ratio of the density of the substance to the density of water
(ii) Characteristics of pressure in liquid are:
-pressure increases in direct proportion to the depth of the liquid -pressure at any point in a liquid acts equally in all directions
-pressure at all points at the same level within a liquid is the same
-pressure applied to an enclosed liquid is transmitted equally throughout the fluid.
(b) Force on the door F = Ah g = 0.4 x 500 x 1.03 x 103x 10 = 2.06x 106N
(c)(i) Effects of heat other than expansion are:
-Rise in temperature -Change of state
-Change of colour
-Change of electrical resistance
(ii) When a liquid and its vapour are in dynamic equilibrium, the vapour is said to be saturated. The pressure or force exerted on unit area of the container of such a saturated vapour is known as saturated vapour pressure.
(iii) Heat generated by coil = heat gained by water + container.
\(\frac{V^2t}{R}\) = mc\(\theta_{\text{water}}\) + Q\(_c \theta_{\text{container}}\)
\(\frac{220^2 \times 2 \times 2 \times 60}{20}\) = m x 4.2 x 10\(^{3}\) x 60 + 100 x 60
290400 = 252000m + 6000
m = \(\frac{290400 - 6000}{252000}\) = 1.13kg
(a)(i) Relative density is the ratio of the density of the substance to the density of water
(ii) Characteristics of pressure in liquid are:
-pressure increases in direct proportion to the depth of the liquid -pressure at any point in a liquid acts equally in all directions
-pressure at all points at the same level within a liquid is the same
-pressure applied to an enclosed liquid is transmitted equally throughout the fluid.
(b) Force on the door F = Ah g = 0.4 x 500 x 1.03 x 103x 10 = 2.06x 106N
(c)(i) Effects of heat other than expansion are:
-Rise in temperature -Change of state
-Change of colour
-Change of electrical resistance
(ii) When a liquid and its vapour are in dynamic equilibrium, the vapour is said to be saturated. The pressure or force exerted on unit area of the container of such a saturated vapour is known as saturated vapour pressure.
(iii) Heat generated by coil = heat gained by water + container.
\(\frac{V^2t}{R}\) = mc\(\theta_{\text{water}}\) + Q\(_c \theta_{\text{container}}\)
\(\frac{220^2 \times 2 \times 2 \times 60}{20}\) = m x 4.2 x 10\(^{3}\) x 60 + 100 x 60
290400 = 252000m + 6000
m = \(\frac{290400 - 6000}{252000}\) = 1.13kg