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S5 Physics Practice Paper: Heat, Optics and Atomic Physics
S5 Physics: Heat, Optics and Atomic Physics
S5 Physics paper: specific heat, latent heat, refraction, lenses, photoelectric effect, energy levels, half-life and E = mc2. 12 questions, 100 marks, 75 minutes, full answers.
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Paper details Level: S5 | Subject: Physics | Time: 75 minutes | Marks: 100 | Questions: 12 | Answers included
Section A: Heat (Q1-Q4)
Section A
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | 0.50 kg of water is heated from 22.0 °C to 78.0 °C. Given that the specific heat capacity of water is 4 200 J kg-1 K-1, find the energy absorbed by the water. | Temperature rise = 78.0 – 22.0 = 56.0 K (1); E = mc(delta T) (2) = 0.50 x 4 200 x 56.0 (2) = 117 600 J, about 1.18 x 10^5 J (3, unit required). |
| 2 | 0.200 kg of water at 90.0 °C is mixed with 0.300 kg of water at 20.0 °C. Neglecting heat loss, find the final temperature of the mixture. | Let the final temperature be T: 0.200 x 4 200 x (90.0 – T) = 0.300 x 4 200 x (T – 20.0) (3); 18.0 – 0.200T = 0.300T – 6.0 (2); 24.0 = 0.500T, so T = 48.0 °C (3). |
| 3 | 0.050 kg of ice at 0 °C is melted completely and then warmed to 20 °C. The specific latent heat of fusion of ice is 3.34 x 10^5 J kg-1 and the specific heat capacity of water is 4 200 J kg-1 K-1. Find the total energy required. | Energy to melt = 0.050 x 3.34 x 10^5 = 16 700 J (4); energy to warm = 0.050 x 4 200 x 20 = 4 200 J (3); total energy = 16 700 + 4 200 = 20 900 J, about 2.09 x 10^4 J (2). |
| 4 | A heater of power 60 W is immersed in 0.30 kg of water for 300 s and the temperature of the water rises by 12.0 °C. Find the efficiency of this heating process. | Energy gained by water = 0.30 x 4 200 x 12.0 = 15 120 J (4); energy supplied = 60 x 300 = 18 000 J (3); efficiency = 15 120 / 18 000 x 100 % = 84.0 % (2). |
Section B: Optics (Q5-Q8)
Section B
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | A ray of light passes from air into glass of refractive index 1.50 at an angle of incidence of 30.0 °. Find the angle of refraction. | By Snell law n = sin i / sin r (2): sin r = sin 30.0 ° / 1.50 = 0.333 (3); hence r = 19.5 ° (3, must be smaller than i). |
| 6 | Find the critical angle of the glass in the previous question. | At the critical angle sin C = 1 / n (3); sin C = 1 / 1.50 = 0.667 (2); C = 41.8 ° (3). |
| 7 | An object 2.0 cm high is placed 15.0 cm in front of a convex lens of focal length 10.0 cm. Find the image distance and the magnification, and state the nature of the image. | 1/v = 1/f – 1/u = 1/10.0 – 1/15.0 = 1/30.0 (3), so v = 30.0 cm (2); magnification m = v/u = 2.0 (2); image 4.0 cm high, inverted and real (2). |
| 8 | Light of frequency 5.0 x 10^14 Hz travels in vacuum. Find its wavelength (c = 3.0 x 10^8 m s-1). | wavelength = c / f (2) = 3.0 x 10^8 / 5.0 x 10^14 (3) = 6.0 x 10^-7 m = 600 nm (2). |
Section C: Atomic Physics (Q9-Q12)
Section C
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | Light of frequency 8.0 x 10^14 Hz is incident on a metal surface of work function 2.20 eV. Given h = 6.63 x 10^-34 J s and 1 eV = 1.60 x 10^-19 J, find the photon energy in eV and the maximum kinetic energy of the emitted photoelectrons. | Photon energy E = hf = 6.63 x 10^-34 x 8.0 x 10^14 = 5.30 x 10^-19 J (2) = 3.32 eV (2); maximum kinetic energy = 3.32 – 2.20 = 1.12 eV (3); positive value confirms the photoelectric effect occurs (2). |
| 10 | An electron in a hydrogen atom makes a transition between two levels whose energy difference is 3.20 x 10^-19 J. Find the frequency and the wavelength of the emitted photon. | E = hf gives f = 3.20 x 10^-19 / 6.63 x 10^-34 = 4.83 x 10^14 Hz (4); wavelength = c / f = 3.0 x 10^8 / 4.83 x 10^14 = 6.21 x 10^-7 m, about 621 nm (4). |
| 11 | A radioactive nuclide has a half-life of 5.0 days. A sample contains 80 mg of the nuclide. Find the mass remaining after 10 days and the time taken for the mass to fall to 10 mg. | 10 days is 2 half-lives, so the mass is 80 / 2^2 = 20 mg (4); falling from 80 mg to 10 mg needs 3 half-lives, i.e. 3 x 5.0 = 15 days (4). |
| 12 | In a nuclear reaction the mass defect is 3.0 x 10^-29 kg. Find the energy released (c = 3.0 x 10^8 m s-1). | E = mc^2 (3) = 3.0 x 10^-29 x (3.0 x 10^8)^2 (3) = 2.7 x 10^-12 J (3, unit required). |
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