AQA A2 Physics P19 Thermal Physics Kerboodle Answers

Banner 1

C19.1 Internal energy and temperature AQA A2 Physics P19 Therma Physics Kerboodle Answers: Page No. 309

1 a. Working of an electrical appliance involves energy transfer processes and one way in which energy is transferred is from mechanical energy to heat energy. So, some amount of energy is lost as heat energy and dissipated into the surroundings. This is why appliances like motors, fans get war after continued usage. 2 a Internal energy is the sum of potential energy of the system and the system’s kinetic energy. The change in internal energy (ΔU) of a reaction is equal to the heat gained or lost (enthalpy change) in a reaction when the reaction is run at constant pressure. 3 a The atoms in a solid are so attracted to each other that they vibrate and don’t move past each other. The molecules of a liquid are attracted to each other, but move more freely and past one another. b When a solid is heated the particles gain energy and start to vibrate faster and faster. Initially the structure is gradually weakened which has the effect of expanding the solid. Further heating provides more energy until the particles start to break free of the structure. Although the particles are still loosely connected they are able to move around. At this point the solid is melting to form a liquid. The particles in the liquid are the same as in the solid but they have more energy. 4 a i the temperature of pure melting ice, 273K ii 20⁰C, 293K iii -196⁰c. 77 K b i 328K ii Calculate the pressure of the gas at 100⁰C. 137 kPa

19.2 Specific heat capacity AQA A2 Physics P19 Therma Physics Kerboodle Answers : Page No.  312

1 23kJ b 535kJ 2 a 280s b 10.3MJ 3 a 320J b 130Jkg^-1 K^-1 4 3.2kW (3.15kW to 3 sig. figs)

19.3 Change of state AQA A2 Physics P19 Therma Physics Kerboodle Answers: Page No.  315

1 a Heat energy is needed to convert solid into a liquid because heat energy increases the kinetic energy of the particles. As the kinetic energy increases, the movement of the particles increases and eventually the bond or attraction between the particles gets broken and the particles start moving faster. Moving faster cause the state of the matter to change because the particles in solid are rigid but when heat is applied the particles are loosened and start moving freely. So it change into a liquid state. b When water freezes at constant temperature, at some point, equilibrium will be reached where all the atoms will maintain a minimum energy level. What that means is that atoms will now be in a state where only vibrational energy will exist. At this state, the internal energy of the system as a whole will significantly decrease. However, it will stay constant. 2 0.16 kg 3 a 4.2 J s^-1 b 6400 s 4 a 22 J s^-1 b 6.5 kJ Banner 3

Practice questions: Page No. 316-317

1 (a) The temperature of the water drops 0⁰C in 10min.  (i) ΔQ = mc Δθ gives energy lost by water = 0.20 × 4200 × 20 = 1.68 × 10⁴ J (ii) Rate of loss of energy = = 28 J s^-1 (or W) (b) (i). ΔQ = m l gives energy to be lost ΔQ = 0.20 × 3.3 × 105 = 6.60 × 104 J Energy = P t gives 6.60 × 104 = 28 t ∴ time taken t = 2.36 × 103 s (39.3 min) (ii) Relevant assumptions:

• energy continues to be lost to the surroundings at the same constant rate
• as in (a)(i) the temperature of the ice formed does not fall below 0 °C

2 (a Water at 18 °C to water at 0 °C: ΔQ = mcΔθ = 1.5 × 4200 × 18 = 1.13 × 10⁵ J Water at 0 °C to ice at 0 °C: ΔQ = ml = 1.5 × 3.3 × 10⁵ = 4.95 × 10⁵ J So total energy released = (1.13 × 10⁵) + (4.95 × 10⁵) = 6.08 × 10⁵ J (b) First point must be included in the answer, plus any one of the following three:

• The ice has to be supplied with energy for it to melt
• The bucket and contents stay at 0 °C for longer
• The bucket with ice extracts more energy from the cans
• The cans are cooled for longer when the bucket contains ice.

3

• plot a graph of temperature against time. (

Gradient of a graph= = 0.056 (± 0.004) °C s^-1 Gradient determined from a clear triangle drawn over more than half of the length of the line. (c) Power of heater gives 48 = 1.0 × c × 0.056 …from which specific heat capacity c of metal =860 (± 60) J kg–1 K–1 (d) ΔQ = ml gives 48 × 200 = 32 × 10–3 × l From which specific latent heat of fusion of ice l = 3.0 × 105 J kg−1 Suitable assumptions include:

• no energy passes to the ice from the surroundings
• none of the energy from the heater is lost to the surroundings
• the temperature of the ice does not change.

4 Thermal energy gained by water ΔQ = mcΔθ = 0.45 × 4200 × 20 = 3.78 × 104 J (b) (i) Thermal energy lost by copper = 3.78 × 10⁴ J (ii) Fall in temperature Δθ of copper is given by ΔQ = mcΔθ ∴ 3.78 × 10⁴ = 0.12 × 390 × Δθ …from which Δθ = 808 °C (or K) (iii) Temperature of copper whilst in flame = 808 + 35 = 843 °C (or 1116 K) 5 (a) Calculate the kinetic energy of the bicycle and rider. (2 marks) Kinetic energy of bicycling and rider = 3.04 × 10³ J (b) (i) Energy converted to thermal energy = 0.60 × 3040 = 1.82 × 103 J Maximum temperature rise of brake blocks is given by using ΔQ = mc Δθ ∴ 1.82 × 103 = 0.12 × 1200 × Δθ …from which Δθ = 12.6 °C (or K) (ii) None of the thermal energy in the brake blocks passes to the surroundings.

(a) (i) Thermal energy generated in 1 min E = Pt = 800 × 60 = 4.80 × 10⁴ J (ii) Temperature rise of her body in 1 min is given by ΔQ = mcΔθ ∴ 4.80 × 104 = 60 × 3900 × Δθ …from which Δθ = 0.205 °C (or K) (b) Energy lost by perspiration in 1 min E = Pt = 500 × 60 = 3.00 × 10⁴ J Mass of sweat evaporated in 1 min is given by ΔQ = ml ∴ 3.00 × 104 = m × 2.3 × 106 …from which m = 1.30 × 10–2 kg (c) When the runner stops running, her temperature falls because:

• she is not generating as much ther malenergy per second…
• but she is still losing heat at the same rate (or she is still sweating).

7 In a geothermal power station, water is pumped through pipes into an underground region of hot rocks. The thermal energy of the rocks heals the water and turns it to steam at high pressure. The steam then drives a turbine at the surface to produce electricity. (a) Water at 21⁰C is pumped into the hot rocks and steam at 100 OC is produced al a rate of 190kgs^-1. (i) Show that the energy per second transferred from the hot rocks the power station in this process is at least 500 MW. Consider 1 second . . . Water from 21 °C to 100 °C: ΔQ = mcΔθ = 190 × 4200 × 79 = 6.30 × 10⁷ J Water at 100 °C into steam at 100 °C: ΔQ = ml = 190 × 2.3 × 10⁶ = 4.37 × 10⁸ J ∴ energy transferred to water per second = (6.30 × 107) + (4.37 × 108) = 5.00 × 10⁸ J ∴ energy is transferred at a rate of 5.00 × 108 W (which is 500 MW) (ii) Mass of rocks m = ρV = 3200 × 4.0 × 10⁶ = 1.28 × 10¹⁰ kg Energy transfer from rocks in 1 day = 5.00 × 10⁸ × 24 × 3600 = 4.32 × 10¹³ J Fall in temperature of rocks is given by ΔQ = mcΔθ ∴ 4.32 × 10¹³ = 1.28 × 10¹⁰ × 850 × Δθ …from which Δθ = 3.97 °C (or K)

DISCLAIMER

Disclaimer: I have tried by level best to provide the answers and video explanations to the best of my knowledge. All the answers and notes are written by me and if there is any similarity in the content then it is purely coincidental. But this is not an alternative to the textbook. You should cover the specification or the textbook thoroughly. This is the quick revision to help you cover the gist of everything. In case you spot any errors then do let us know and we will rectify it. References: BBC Bitesize AQA GCSE Science Kerboodle textbook Wikipedia Wikimedia Commons Join Our Free Facebook Group : Get A* in GCSE and A LEVEL Science and Maths by Mahima Laroyia: https://www.facebook.com/groups/expertguidance.co.uk/ For Free Tips, advice and Maths and Science Help