c(ii)
(a)(i) Define dew point. (ii) Explain why dew forms more quickly on the metal parts than on the rubber parts of a bicycle placed in the open overnight.
(b)(i) Explain the statement. the specific heat capacity of copper is 400 J/kg/K. (ii) Two metals, P and Q are supplied with the same quantity of heat.
If the ratio of the specific heat capacity of P to Q is 3: 1 and their masses are in the ratio I:2 respectively.
calculate the ratio of the temperature rise of P to Q.
(c)(i) Define coefficient of thermal conductivity of a material.
The diagram above c(ii) illustrates a composite bar of iron and copper. The bar is insulated along its sides and it has a diameter of 10 mm. The length and thermal conductivity of the iron are 0.15 m and 40 W/m/K, respectively and those of copper are 0.05 m and 360 W/m/K, respectively. If the free ends of the iron and copper are kept at 100°C and 0°C respectively. calculate the (i) temperature at the interface between the bars; (ii) rate of heat flow along the bar.
(a)(i) Dew Point: This is the temperature at which the water vapour/miosture present in the atmosphere/air is just sufficient to saturate it.
(a)(ii) Explanation of why dew forms more quickly on metal parts than rub Metal is a better conductor of heat than rubber. At night. metal loses heat faster than rubber and becomes colder.
Warm air gains latent heat of vaporization from the faster than rubber. Dews are thus formed on the metal parts.
9: (b)(i) Explanation of the statement: 400J of heat/energy is required to charge or raise the temperature of lkg of copper by 1k.
(ii) Heat = mcΔΘ
Hence; MpCpΔΘP = MqCqΔΘq
Given that: Cp : Cq = 3:1 and
Mp : Mq = 1:2
\(\frac{ΔΘp}{ΔΘq}\) = \(\frac{Cp * Mq}{Cp * Mp}\) → \(\frac{1}{3}\) * \(\frac{2}{1}\)
ΔΘp : ΔΘq = 2:3
Heat = mcΔΘ
Hence; MpCpΔΘP = MqCqΔΘq
1 * 3 * ΔΘp = 2 * 1 * ΔΘp
\(\frac{ ΔΘp }{ ΔΘq}\) = \(\frac{2}{3}\)
ΔΘp : ΔΘq = 2:3
(c) Definition of Coefficient of Thermal Conductivity: The time rate of flow of heat per unit cross sectional area per unit temperature gradient.
(cii) \(\frac{Θ}{t} = \frac{KAΔΘ}{ΔL}\)
\(40 \times π \frac{(10 \times10^{-3})^2}{4}\times\frac{(100 - Θ)}{0.15}\)
= \(360 \times π \frac{(10 \times10^{-3})^2}{4} \times\frac {Θ - 0}{0.05}\) = 3.6°C
(c)(iii) \(\frac{Θ}{t} = 40π (\frac{10^{-2}}{4})^3 (\frac{100 - 3.6}{0.15}) = 2.02J/S\)
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