Course: Combined Science + Separate Physics | Tier: Foundation + Higher | Equation sheet: given in the exam
Equations used: P = VI | P = I²R | E = Pt | Q = It | E = QV | ΔE = mcΔθ | E = mL | efficiency
Key idea: the energy a heater transfers (E = Pt or E = QV) is the energy that warms the substance (ΔE = mcΔθ) or changes its state (E = mL). Temperature changes use c; changes of state use L – the temperature stays the same while it melts or boils.
Data: c water = 4200 J/kg °C | L fusion of ice = 3.34 × 105 J/kg | L vaporisation of water = 2.26 × 106 J/kg. Assume no energy is wasted unless told otherwise.
Watch the units in Q6–10: convert times to seconds (minutes × 60, hours × 3600), masses to kg (g ÷ 1000) and power to W (kW × 1000). 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) A 2000 W kettle is switched on for 168 s. It heats 1.0 kg of water. Calculate the energy transferred, then the temperature rise of the water.
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Step 1 – energy transferred
F: E = P t
I: E = 2000 × 168
A: E = 336 000 J
Step 2 – temperature rise
F: ΔE = m c Δθ
I: 336 000 = 1.0 × 4200 × Δθ
F: Δθ = 336 000 ÷ 4200
A: 80 °C
Q2 (F) A 1.5 kW immersion heater is switched on for 280 s in a tank holding 2.5 kg of water. Calculate the temperature rise.
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Step 1 – energy transferred (1.5 kW = 1500 W)
F: E = P t
I: E = 1500 × 280
A: E = 420 000 J
Step 2 – temperature rise
F: ΔE = m c Δθ
I: 420 000 = 2.5 × 4200 × Δθ, so 420 000 = 10 500 × Δθ
F: Δθ = 420 000 ÷ 10 500
A: 40 °C
Q3 (F) A kettle on the 230 V mains draws a current of 10 A. The water is already boiling. How long does it take to boil away 0.23 kg of water?
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Step 1 – power
F: P = V I
I: P = 230 × 10
A: P = 2300 W
Step 2 – energy to boil away the water
F: E = m L
I: E = 0.23 × 2 260 000
A: E = 519 800 J
Step 3 – time
F: E = P t
I: 519 800 = 2300 × t
F: t = 519 800 ÷ 2300
A: 226 s
Q4 (F) A 12 V heater draws 4.0 A. It is used to melt 0.12 kg of ice at 0 °C. How long does this take?
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Step 1 – power
F: P = V I
I: P = 12 × 4.0
A: P = 48 W
Step 2 – energy to melt the ice
F: E = m L
I: E = 0.12 × 334 000
A: E = 40 080 J
Step 3 – time
F: E = P t
I: 40 080 = 48 × t
F: t = 40 080 ÷ 48
A: 835 s (about 14 minutes)
Q5 (F) 700 C of charge flows through a 12 V heater placed in 0.20 kg of water. Calculate the temperature rise of the water.
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Step 1 – energy transferred
F: E = Q V
I: E = 700 × 12
A: E = 8400 J
Step 2 – temperature rise
F: ΔE = m c Δθ
I: 8400 = 0.20 × 4200 × Δθ, so 8400 = 840 × Δθ
F: Δθ = 8400 ÷ 840
A: 10 °C
Q6 (F/H) A 2.5 kW kettle is switched on for 2 minutes 40 seconds. It heats the water inside from 20 °C to 100 °C. Calculate the mass of water in the kettle, to 2 significant figures.
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Convert first: 2.5 kW = 2500 W; 2 min 40 s = (2 × 60) + 40 = 160 s
Step 1 – energy transferred
F: E = P t
I: E = 2500 × 160
A: E = 400 000 J
Step 2 – mass
F: ΔE = m c Δθ
I: 400 000 = m × 4200 × 80, so 400 000 = 336 000 × m
F: m = 400 000 ÷ 336 000 = 1.19… kg
A: 1.2 kg (2 s.f.)
Q7 (F/H) A heating element with a resistance of 20 Ω carries a current of 10 A for 8 minutes 21 seconds. It is used to melt ice at 0 °C. What mass of ice melts?
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Convert first: 8 min 21 s = (8 × 60) + 21 = 501 s
Step 1 – power
F: P = I² R
I: P = 10² × 20
A: P = 2000 W
Step 2 – energy
F: E = P t
I: E = 2000 × 501
A: E = 1 002 000 J
Step 3 – mass melted
F: E = m L
I: 1 002 000 = m × 334 000
F: m = 1 002 000 ÷ 334 000
A: 3.0 kg
Q8 (F/H) An ice rink is made by freezing 1.5 × 105 kg of water that is already at 0 °C. The freezing machine removes energy at a rate of 2.0 × 105 W. Calculate the energy that must be removed, in standard form, then how many hours the freezing takes, to 2 significant figures.
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Step 1 – energy removed (freezing releases the same latent heat as melting takes in)
F: E = m L
I: E = 1.5 × 105 × 3.34 × 105
A: E = 5.01 × 1010 J
Step 2 – time
F: E = P t
I: 5.01 × 1010 = 2.0 × 105 × t
F: t = 5.01 × 1010 ÷ 2.0 × 105 = 2.505 × 105 s
A: 2.505 × 105 ÷ 3600 = 69.6 hours = 70 hours (2 s.f.)
Q9 (H) A kettle on the 230 V mains draws 10 A. It takes 3 minutes 20 seconds to heat 1200 g of water from 20 °C to 100 °C. Calculate the efficiency of the kettle as a percentage, to 2 significant figures.
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Convert first: 3 min 20 s = (3 × 60) + 20 = 200 s; 1200 g = 1.2 kg
Step 1 – power: P = V I = 230 × 10 = 2300 W
Step 2 – total energy input
F: E = P t
I: E = 2300 × 200
A: E = 460 000 J
Step 3 – useful energy
F: ΔE = m c Δθ
I: ΔE = 1.2 × 4200 × 80
A: ΔE = 403 200 J
Step 4 – efficiency
F: efficiency = useful output ÷ total input
I: efficiency = 403 200 ÷ 460 000 = 0.876…
A: 88% (2 s.f.)
Q10 (H) A 2.0 kW kettle that is 90% efficient contains 100 g of water at 20 °C. How many minutes does it take to heat the water to 100 °C and boil it all away? Give your answer to 2 significant figures.
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Convert first: 100 g = 0.10 kg; 2.0 kW = 2000 W
Step 1 – heating to 100 °C
F: ΔE = m c Δθ
I: ΔE = 0.10 × 4200 × 80
A: ΔE = 33 600 J
Step 2 – boiling away
F: E = m L
I: E = 0.10 × 2.26 × 106
A: E = 226 000 J
Step 3 – total input needed (useful = 33 600 + 226 000 = 259 600 J)
F: efficiency = useful output ÷ total input
I: 0.90 = 259 600 ÷ input
F: input = 259 600 ÷ 0.90 = 288 444 J
Step 4 – time
F: E = P t
I: 288 444 = 2000 × t
F: t = 288 444 ÷ 2000 = 144.2 s
A: 144.2 ÷ 60 = 2.40… = 2.4 minutes (2 s.f.)
