1. (a) Plot a graph with time/s on the x-axis and temp/°C on the y-axis using the provided data points: (0, 20.1), (60, 23.0), (120, 26.9), (180, 30.0), (240, 33.1), (300, 36.9). Draw a best-fit straight line through the points.
(b) Gradient = (change in temperature) / (change in time) = (36.9 - 20.1) °C / (300 - 0) s = 16.8 / 300 = 0.056 °C/s
(c) Power P = 48 W, mass m = 1.0 kg, gradient dT/dt = 0.056 °C/s.
Using P = mc(dT/dt), so c = P / (m * dT/dt) = 48 / (1.0 * 0.056) = 857 J/kg°C (or 860 J/kg°C to 2 significant figures).
(d) Energy supplied by heater in 200 s: E = P * t = 48 W * 200 s = 9600 J.
This energy melts 32 g = 0.032 kg of ice.
Assuming all energy goes into melting ice (no heat loss, ice at 0°C), latent heat L = E / m = 9600 J / 0.032 kg = 300,000 J/kg = 3.0 × 10⁵ J/kg.
2. (a) Mass of water heated: m = 750 g - 25 g = 725 g = 0.725 kg.
Temperature change: ΔT = 100°C - 20°C = 80°C.
Energy supplied: E = P * t = 2000 W * 120 s = 240,000 J.
Using E = mcΔT, so c = E / (mΔT) = 240000 / (0.725 * 80) = 240000 / 58 = 4138 J/kg°C ≈ 4140 J/kg°C.
(b) (i) Mass of water boiled away: m_vap = 94 g = 0.094 kg.
Energy used for vaporization: E_vap = P * t = 2000 W * 105 s = 210,000 J.
Latent heat of vaporization L_v = E_vap / m_vap = 210000 / 0.094 = 2,234,043 J/kg ≈ 2.23 × 10⁶ J/kg.
(ii) Assumptions: All energy from the heater goes into vaporizing water; the beaker does not absorb or lose significant heat during boiling; the water is at 100°C when boiling starts.
3. (a) (i) Energy lost = mcΔT = 0.20 kg * 4200 J/kg°C * (20 - 0)°C = 0.20 * 4200 * 20 = 16,800 J.
(ii) Average rate of energy loss = Energy / time = 16800 J / (10 * 60 s) = 16800 / 600 = 28 J/s.
(b) (i) Energy needed to freeze: E_freeze = mL_f = 0.20 kg * 3.3 × 10⁵ J/kg = 66,000 J.
Time to freeze = E_freeze / rate = 66000 J / 28 J/s = 2357 s ≈ 39.3 minutes.
(ii) Assumption: The average rate of energy loss (28 J/s) remains constant while the water freezes.
Parent Tip: Review the logic above to help your child master the concept of latent heat worksheet.
