Physics · May / June 2023 · Paper 41 · Question 4
0625
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The question as text
A student investigates the efficiency of a filament lamp. Fig. 4.1 shows the filament lamp with its glass bulb immersed in water in a beaker. filament lamp to power supply water beaker thermometer Fig. 4.1 The reading on the thermometer in the water is 19.0 °C. Only the glass of the lamp is in contact with the water and the electrical connections are completely insulated. The lamp is switched on. At the end of the experiment, the temperature of the water is 21.5 °C.
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4(a)(i)[3]
c = (∆)E / m∆θ OR (∆E =) mc∆θ
(∆θ =) 21.5 – 19 OR (∆θ =) 2.5 (°C)
(∆E =) 0.6(0) × 4200 × 2.5 OR (∆E =) 0.6(0) × 4200 × {21.5 – 19}
4(a)(ii)[4]
(maximum possible efficiency =) 3.1% or 0.031
E = Pt OR (E =) Pt OR (E =) 13 × 500 OR (E =) 6500
C1 (useful energy output =) 6500 – 6300 OR (useful energy output =) 200
C1 efficiency = useful energy (output) / total energy (input) (× 100%)
C1 OR (efficiency =) useful energy (output) / total energy (input) (× 100%)
OR (efficiency =) {6500 – 6300} / 6500
OR (efficiency =) 200 / 6500 (× 100%)
OR
P = E/t OR (P =) E / t OR (P =) 6 300 / 500 OR (P =) 12.6 (W)
(C1) (useful power output =) total power (output) – wasted power (output)
(C1) OR (useful power output =) 13 – {6300 / 500}
OR (useful power output =) 13–12.6
efficiency = useful power (output) / total power (input) (× 100%)
(C1) OR (efficiency =) useful power (output) / total power (input) (× 100%)
OR (efficiency =) 0.4 / 13 (× 100%)
4(b)[1]
any one from:
temperature change is an underestimate (due to thermal energy losses)
(thermal energy is) transferred from the water (to air / beaker / bench)
energy (other than light) transferred in lamp (filament / glass / internal structure)
(some) water evaporates
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