For each question: list what you know, decide which equation to use, then use FIFA. Key choice in this topic: if the temperature changes, use ΔE = mcΔθ; if the state changes (melting, boiling) at a constant temperature, use E = mL. Data: c water = 4200 J/kg °C; L fusion of ice = 334 000 J/kg; L vaporisation of water = 2.26 × 106 J/kg.
Q1 (F) A 2.7 kg block has a volume of 0.0010 m³. Calculate its density.
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Which equation? Know mass and volume → ρ = m / V
I: ρ = 2.7 ÷ 0.0010
A: 2700 kg/m³
Q2 (F) How much energy is needed to melt 0.20 kg of ice at 0 °C?
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Which equation? Change of state, no temperature change → E = mL
I: E = 0.20 × 334 000
A: 66 800 J
Q3 (F) 2.0 kg of water is heated from 20 °C to 70 °C. How much energy is needed?
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Which equation? Temperature changes → ΔE = mcΔθ
I: ΔE = 2.0 × 4200 × 50
A: 420 000 J
Q4 (F) Oil has a density of 920 kg/m³. What is the mass of 0.50 m³ of oil?
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Which equation? Know density and volume → ρ = m / V
I: 920 = m ÷ 0.50
F: m = 920 × 0.50
A: 460 kg
Q5 (F/H) 1.13 × 106 J of energy boils away 0.50 kg of a liquid at its boiling point. Calculate its specific latent heat of vaporisation.
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Which equation? Change of state → E = mL
I: 1.13 × 106 = 0.50 × L
F: L = 1.13 × 106 ÷ 0.50
A: 2.26 × 106 J/kg
Q6 (F/H, S only) A gas at 100 kPa has a volume of 2.0 m³. It is compressed to 0.50 m³ at constant temperature. Calculate the new pressure.
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Which equation? Gas, constant temperature, pressure and volume change → pV = constant (p₁V₁ = p₂V₂)
I: 100 × 2.0 = p₂ × 0.50
F: p₂ = 200 ÷ 0.50
A: 400 kPa
Q7 (F/H) A stone has a mass of 250 g and a volume of 100 cm³. Calculate its density in kg/m³.
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Which equation? → ρ = m / V
I: ρ = 250 ÷ 100 = 2.5 g/cm³
A: × 1000 = 2500 kg/m³
Q8 (F/H) 84 000 J of energy raises the temperature of 2.0 kg of a liquid by 10 °C. Calculate the specific heat capacity.
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Which equation? Temperature change → ΔE = mcΔθ
I: 84 000 = 2.0 × c × 10
F: c = 84 000 ÷ 20
A: 4200 J/kg °C
Q9 (H) A block of ice at 0 °C has a volume of 0.020 m³ (density of ice = 920 kg/m³). How much energy is needed to melt it? Give your answer in standard form to 2 s.f.
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Which equations? Need mass first → ρ = m / V, then melting → E = mL
Step 1: m = 920 × 0.020 = 18.4 kg
Step 2: E = 18.4 × 334 000 = 6 145 600 J
A: 6.1 × 106 J
Q10 (H, S only) A syringe holds 60 cm³ of air at 100 kPa. The plunger is pushed in slowly until the pressure is 150 kPa. What is the new volume?
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Which equation? Slowly = constant temperature → pV = constant
I: 100 × 60 = 150 × V₂
F: V₂ = 6000 ÷ 150
A: 40 cm³ (units match, so no conversion needed)
Q11 (H) Water at 100 °C is heated and 3.0 × 105 J of energy is supplied. What mass of water boils away? Give your answer to 2 s.f.
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Which equation? Already at 100 °C, so it’s a change of state → E = mL
I: 3.0 × 105 = m × 2.26 × 106
F: m = 3.0 × 105 ÷ 2.26 × 106 = 0.1327…
A: 0.13 kg
Q12 (H) A 2.0 kW heater is placed in ice at 0 °C for 60 s. What mass of ice melts? Give your answer to 2 s.f.
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Which equations? Energy from the heater → E = Pt; melting → E = mL
Step 1: E = 2000 × 60 = 120 000 J
Step 2: 120 000 = m × 334 000, so m = 0.359…
A: 0.36 kg
