Physics

1. 1.

   A 0.250 kg aluminium block warms from 20.0 to 60.0 degrees C. Specific heat capacity is 900 J kg^-1 K^-1. Find energy transferred, assuming no losses, to 3 significant figures.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Temperature differences have the same numerical size in kelvin and Celsius.
   2. Only the difference belongs in mc delta T; multiplying by the final temperature would overestimate energy.

   Marking points

   • Temperature change is 40.0 K.
   • Q = mc delta T.
   • Q = 9.00e3 J.

   Examiner tip: Do not add 273 to a temperature change.

2. 2.

   A fixed amount of ideal gas in a sealed rigid container is initially at 100 kPa and 300 K. It is heated at fixed volume to 450 K with no gas entering or leaving. Find final pressure and explain the molecular reason.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. From pV = nRT, all factors except p and T remain fixed.
   2. Both more frequent impacts and greater momentum changes contribute to greater pressure in the ideal kinetic model.

   Marking points

   • At fixed amount and volume, p/T is constant.
   • Final pressure is 150 kPa.
   • Higher mean molecular kinetic energy increases momentum transfer to walls per unit time.

   Examiner tip: Use absolute temperature in gas ratios.

3. 3.

   Mix 0.200 kg water at 80.0 degrees C with 0.300 kg water at 20.0 degrees C. Find equilibrium temperature. Neglect heat loss and container heat capacity; specific heat is identical and constant.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The final temperature is a mass-weighted average, not an arithmetic mean of the temperatures.
   2. More cold water shifts the result below 50 degrees C; both samples reach the same final temperature.

   Marking points

   • Heat lost equals heat gained: 0.200c(80.0 - T) = 0.300c(T - 20.0).
   • Cancel c and solve 22.0 = 0.500T.
   • T = 44.0 degrees C.

   Examiner tip: Container heat capacity would change the balance unless it were negligible as stated.

4. 4.

   Find rms molecular speed for an ideal gas at 300 K with molar mass 0.0280 kg mol^-1, using R = 8.31 J mol^-1 K^-1. Explain why this is not the mean velocity. Give 3 significant figures.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The molar form pairs R with mass per mole, not mass per molecule.
   2. Squaring speed before averaging prevents cancellation of opposite directions.

   Marking points

   • v_rms = sqrt(3RT/M).
   • v_rms = 517 m s^-1.
   • In equilibrium random directions make mean velocity zero, while mean squared speed is positive.

   Examiner tip: Convert grams per mole to kilograms per mole before using SI R.

5. 5.

   Place 0.0500 kg ice at 0 degrees C in 0.200 kg water at 30.0 degrees C in an insulated container of negligible heat capacity. Use c_water = 4200 J kg^-1 K^-1 and L_f = 3.34e5 J kg^-1. Decide whether all ice melts and find final temperature.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Test the melting energy first; if the warm water could not supply it the final state would contain ice at zero degrees C.
   2. After melting, the original water and melted ice share the surplus 8500 J over a combined mass 0.250 kg.

   Marking points

   • Water can release 25200 J cooling to 0 degrees C; melting needs 16700 J, so all ice melts.
   • 0.200c(30.0 - T) = 0.0500L_f + 0.0500cT.
   • Final temperature is 8.10 degrees C (3 significant figures).

   Examiner tip: Include both latent heat and warming of the melted water.

6. 6.

   Exactly 0.500 mol ideal gas expands reversibly and isothermally at 300 K from 0.0100 to 0.0200 m^3. Given work by gas W = nRT ln(V2/V1), use R = 8.31 J mol^-1 K^-1 to find work and heat supplied. State internal energy change and the sign convention.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The logarithmic work follows from integrating p = nRT/V during quasistatic expansion.
   2. Heat must replace the energy leaving as work to keep the ideal gas temperature constant.

   Marking points

   • W = 0.500*8.31*300*ln(2) = 864 J (3 significant figures), positive for expansion.
   • Ideal-gas internal energy depends only on temperature, so delta U = 0 J.
   • Using delta U = Q - W with Q into gas positive, heat supplied Q = 864 J.

   Examiner tip: Isothermal does not mean no heat transfer; it means constant temperature.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.