Preview Cambridge International AS and A Level Physics Workbook

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Cambridge International AS and A level Physics

c A resistance thermometer may be based on a resistance wire or a thermistor. State which of these two types of thermometer shows a big change in resistance over a small change in temperature. d A resistance thermometer is likely to have a greater mass than a thermocouple. Explain why this means that a thermocouple thermometer is likely to respond more quickly to a change in temperature. e A thermocouple has a non-linear response to temperature and hence will require calibration if it is to be used as a thermometer. Explain the meanings of the terms non-linear and calibration. 4 The thermodynamic (Kelvin) scale of temperature has two fixed points. a State the lower fixed point and give its value in K and in °C. b The upper fixed point is the triple point of water. Explain what this means. c Give the value of the upper fixed point in K and in °C.

Exercise 21.3 Energy change calculations: s.h.c. and s.l.h. Knowing the values of a material’s specific heat capacity and its specific latent heat allows us to calculate energy changes. This is an exercise in calculating and using both of these quantities. 140

1 The equation c =

E m Δθ

defines specific heat capacity.

a State what each symbol represents, and give its SI unit. (Note: Δθ is a single quantity.) The specific heat capacity of lead is 126 J kg −1 K−1. b Calculate the amount of energy required to raise the temperature of 1 kg of lead by 10 °C. c 1 kJ of energy is supplied to 1 kg of lead. Determine the amount by which the temperature of the lead will rise, to the nearest 1 °C. 2 To cool a 5 kg block of steel at a temperature of 200 °C, it is plunged into a tank containing 50 kg of water at 20 °C. You can calculate the final temperature of the water like this: The steel and the water will reach the same final temperature. We can write this as final temperature = X °C. specific heat capacity of steel = 450 J kg −1 K−1 specific heat capacity of water = 4200 J kg −1 K−1 a Write down an expression in terms of X for the energy lost by the steel in cooling from 200 °C to X. b Write down a similar expression for the energy gained by the water in being heated from 20 °C to X. c These two quantities are equal (assuming no energy is lost to the surroundings). Write an equation in which these quantities are shown to be equal. d Solve the equation for X. e You could use a method like this to determine the specific heat capacity of a material. Outline how you would do this. List the quantities you would measure. State one other quantity whose value you would need to know.


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