A-Level Internal energy, heating and work done, the laws of thermodynamics, and specific heat and latent heat

(a) Internal energy is the sum of the random kinetic energies and the potential energies of all the molecules/particles in the coin. The kinetic part is due to the random motion (vibration) of the particles; the potential part is due to the forces (bonds) between the particles. (b) Loss of GPE = mgh = 0.00500 x 9.81 x 40.0 = 1.962 J. Energy raising internal energy = 10.0% = 0.1962 J. Delta U = m c Delta T => Delta T = 0.1962/(0.00500 x 385) = 0.1962/1.925 = 0.102 K. (c) For the coin dropping and landing: - dU (internal energy increase): POSITIVE (temperature rises). - q (heat added): ZERO (the impact is rapid; essentially no heat is transferred to the coin). - w (work done ON the coin): POSITIVE (the ground does work on the coin as it is brought to rest/deforms). Check: dU = q + w gives positive = 0 + positive.

Principle: molar heat capacity C is defined by Q = n C ΔT, that is, the heat needed to raise one mole by one kelvin. To use it we need three things: the number of moles n, the heat supplied Q, and the temperature rise ΔT. Only ΔT is given directly, so the other two must be built up. Step 1 - find the number of moles from the number of molecules. One mole contains the Avogadro number of molecules, N_A = 6.02 × 10²³ mol⁻¹, so n = N / N_A n = 1.41 × 10²⁴ / 6.02 × 10²³ n = 2.34 mol Step 2 - find the heat supplied, using the first law of thermodynamics. The first law says the heat supplied to a gas either raises its internal energy or is used by the gas to do work on its surroundings as it expands: Q = ΔU + W Here ΔU = +2480 J (the internal energy climbs) and W = +1760 J (the gas does work on its surroundings, which it must, because the pressure is fixed so the gas expands as it warms). Q = 2480 J + 1760 J Q = 4240 J Step 3 - find the temperature rise. ΔT = 150 °C − 65 °C = 85 °C A temperature DIFFERENCE has the same numerical value on the Celsius and kelvin scales, because the two scales have the same size degree, so ΔT = 85 K (There is no need to convert each temperature to kelvin separately here.) Step 4 - rearrange Q = n C ΔT for C by dividing both sides by n ΔT. C = Q / (n ΔT) Step 5 - substitute. n ΔT = 2.34 mol × 85 K = 199 mol K C = 4240 J / 199 mol K C = 21.3 J mol⁻¹ K⁻¹ The molar heat capacity at constant pressure is about 21 J mol⁻¹ K⁻¹.

Principle: the first law of thermodynamics, ΔU = q + w, where q is the thermal energy supplied TO the gas, w is the work done ON the gas, and ΔU is the increase in internal energy. Signs matter: q is positive when heat enters the gas, w is positive when the surroundings do work on the gas (compression) and negative when the gas does work on the surroundings (expansion). Row A → B. Heat enters the gas, so q = +1370 J. The work has magnitude 390 J. Heat entering while the gas pushes outwards means the gas does work on its surroundings, i.e. the work done ON the gas is negative: w = −390 J. ΔU = q + w = 1370 + (−390) = +980 J. Row B → C. 'No heat is exchanged' means the change is adiabatic, so q = 0. 550 J of work is done ON the gas, so w = +550 J. ΔU = q + w = 0 + 550 = +550 J. Row C → A. Step 1 - use the fact that internal energy is a function of state. A → B → C → A returns the gas to exactly the state it started in, so its internal energy returns to its starting value and the changes must sum to zero round the complete cycle: ΔU(A→B) + ΔU(B→C) + ΔU(C→A) = 0 980 + 550 + ΔU(C→A) = 0 ΔU(C→A) = −1530 J. Step 2 - the work. From part (a), C → A takes place at constant volume, and no change in volume means the gas neither pushes back the surroundings nor is pushed in, so w = 0. Step 3 - rearrange the first law for q by subtracting w from both sides: q = ΔU − w q = −1530 − 0 q = −1530 J. The negative sign means 1530 J of thermal energy leaves the gas over C → A. Completed Table 3.1: A → B: q = +1370 J, w = −390 J, ΔU = +980 J B → C: q = 0, w = +550 J, ΔU = +550 J C → A: q = −1530 J, w = 0, ΔU = −1530 J