Course: Combined Science + Separate Physics | Tier: Foundation + Higher | Equation sheet: given in the exam
Equations used: ρ = m/V | W = mg | p = F/A | Eₚ = mgh | ΔE = mcΔθ | E = mL | P = E/t
Key idea: questions often give you a volume (or the dimensions of an object) when the next equation needs a mass. Use density to get the mass first: m = ρ × V.
Data: water ρ = 1000 kg/m³, c = 4200 J/kg °C | ice ρ = 920 kg/m³, L fusion = 3.34 × 105 J/kg | g = 9.8 N/kg
Watch the units in Q6–10: 1 litre = 0.001 m³ | 1 cm³ = 1 × 10−6 m³ | 1 cm² = 1 × 10−4 m² | cm ÷ 100 = m | minutes × 60 = s. Some answers need standard form.
Foundation: aim for Q1–8. Higher: aim for Q4–10. Use FIFA for every step and don’t round until the end.
Questions
Q1 (F) An aluminium block has a volume of 0.002 m³. Aluminium has a density of 2700 kg/m³. Calculate the mass of the block, then its weight.
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Step 1 – mass
F: ρ = m / V
I: 2700 = m ÷ 0.002
F: m = 2700 × 0.002
A: m = 5.4 kg
Step 2 – weight
F: W = m g
I: W = 5.4 × 9.8
A: 52.9 N (3 s.f.)
Q2 (F) A fish tank measures 0.50 m × 0.30 m × 0.40 m and is full of water. Calculate the weight of the water.
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Step 1 – volume: 0.50 × 0.30 × 0.40 = 0.06 m³
Step 2 – mass
F: ρ = m / V
I: 1000 = m ÷ 0.06
F: m = 1000 × 0.06
A: m = 60 kg
Step 3 – weight
F: W = m g
I: W = 60 × 9.8
A: 588 N
Q3 (F) A concrete block measures 0.40 m × 0.20 m × 0.10 m. Concrete has a density of 2400 kg/m³. The block rests on its 0.40 m × 0.20 m face. Calculate the pressure it exerts on the ground.
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Step 1 – volume and mass
V = 0.40 × 0.20 × 0.10 = 0.008 m³
F: ρ = m / V
I: 2400 = m ÷ 0.008
F: m = 2400 × 0.008 = 19.2 kg
Step 2 – weight: W = m g = 19.2 × 9.8 = 188.16 N
Step 3 – pressure (area = 0.40 × 0.20 = 0.08 m²)
F: p = F / A
I: p = 188.16 ÷ 0.08
A: 2352 Pa (≈ 2400 Pa)
Q4 (F) A paddling pool holds 1.5 m³ of water. The Sun warms the water by 3 °C. Calculate the energy the water absorbs, in MJ.
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Step 1 – mass
F: ρ = m / V
I: 1000 = m ÷ 1.5
F: m = 1000 × 1.5
A: m = 1500 kg
Step 2 – energy
F: ΔE = m c Δθ
I: ΔE = 1500 × 4200 × 3
A: ΔE = 18 900 000 J = 18.9 MJ
Q5 (F) A tray of ice cubes at 0 °C has a total volume of 250 cm³. Calculate the mass of the ice, then the energy needed to melt it.
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Convert first: 250 cm³ = 2.5 × 10−4 m³
Step 1 – mass
F: ρ = m / V
I: 920 = m ÷ 2.5 × 10−4
F: m = 920 × 2.5 × 10−4
A: m = 0.23 kg
Step 2 – energy to melt
F: E = m L
I: E = 0.23 × 334 000
A: 76 820 J (≈ 77 kJ)
Q6 (F/H) A pump lifts 600 litres of water into a water tower 15 m high in 2.5 minutes. Calculate the useful power of the pump.
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Convert first: 600 litres = 600 × 0.001 = 0.60 m³; 2.5 minutes = 150 s
Step 1 – mass
F: ρ = m / V
I: 1000 = m ÷ 0.60
F: m = 1000 × 0.60 = 600 kg
Step 2 – GPE gained
F: Eₚ = m g h
I: Eₚ = 600 × 9.8 × 15
A: Eₚ = 88 200 J
Step 3 – power
F: P = E / t
I: P = 88 200 ÷ 150
A: 588 W
Q7 (F/H) A 2.0 kW deep-fat fryer holds 2.0 litres of oil (ρ = 920 kg/m³, c = 2000 J/kg °C). How many minutes does it take to heat the oil from 20 °C to 180 °C? Give your answer to 2 significant figures.
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Convert first: 2.0 litres = 0.0020 m³; 2.0 kW = 2000 W
Step 1 – mass
F: ρ = m / V
I: 920 = m ÷ 0.0020
F: m = 920 × 0.0020 = 1.84 kg
Step 2 – energy
F: ΔE = m c Δθ
I: ΔE = 1.84 × 2000 × 160
A: ΔE = 588 800 J
Step 3 – time
F: P = E / t
I: 2000 = 588 800 ÷ t
F: t = 588 800 ÷ 2000 = 294.4 s
A: 294.4 ÷ 60 = 4.90… = 4.9 minutes (2 s.f.)
Q8 (F/H) A gold bar measures 25 cm × 8.0 cm × 3.2 cm. Gold has a density of 1.93 × 104 kg/m³. Calculate the weight of the bar, to 3 significant figures.
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Step 1 – volume: 25 × 8.0 × 3.2 = 640 cm³ = 640 × 10−6 = 6.4 × 10−4 m³
Step 2 – mass
F: ρ = m / V
I: 1.93 × 104 = m ÷ 6.4 × 10−4
F: m = 1.93 × 104 × 6.4 × 10−4
A: m = 12.352 kg
Step 3 – weight
F: W = m g
I: W = 12.352 × 9.8 = 121.05 N
A: 121 N (3 s.f.)
Watch out: 1 cm³ is 10−6 m³, not 10−2 m³.
Q9 (H) A hot water tank holds 5.0 × 104 cm³ of water. A 3.0 kW immersion heater warms it from 15 °C to 60 °C. How many minutes does this take? (Assume no energy is wasted.)
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Convert first: 5.0 × 104 cm³ = 5.0 × 104 × 10−6 = 5.0 × 10−2 m³ (0.050 m³); 3.0 kW = 3000 W
Step 1 – mass
F: ρ = m / V
I: 1000 = m ÷ 0.050
F: m = 1000 × 0.050 = 50 kg
Step 2 – energy
F: ΔE = m c Δθ
I: ΔE = 50 × 4200 × 45
A: ΔE = 9 450 000 J (9.45 × 106 J)
Step 3 – time
F: P = E / t
I: 3000 = 9.45 × 106 ÷ t
F: t = 9.45 × 106 ÷ 3000 = 3150 s
A: 3150 ÷ 60 = 52.5 minutes
Q10 (H) A patch of ice on a pond has an area of 2.0 × 104 cm² and is 5.0 cm thick. It is at 0 °C. Sunlight delivers 500 W to it. Assuming all this energy melts the ice, how many hours does it take to melt completely? Give your answer to 2 significant figures.
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Convert first: 2.0 × 104 cm² = 2.0 × 104 × 10−4 = 2.0 m²; 5.0 cm = 0.050 m
Step 1 – volume and mass
V = 2.0 × 0.050 = 0.10 m³
F: ρ = m / V
I: 920 = m ÷ 0.10
F: m = 920 × 0.10 = 92 kg
Step 2 – energy to melt
F: E = m L
I: E = 92 × 3.34 × 105
A: E = 3.0728 × 107 J
Step 3 – time
F: P = E / t
I: 500 = 3.0728 × 107 ÷ t
F: t = 3.0728 × 107 ÷ 500 = 61 456 s
A: 61 456 ÷ 3600 = 17.07… = 17 hours (2 s.f.)
