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A student investigates the efficiency of a filament lamp. Fig. 4.1 shows the filament lamp with itsglass bulb immersed in water in a beaker.filament lampto power supplywaterbeakerthermometerFig. 4.1The 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 completelyinsulated.The lamp is switched on.At the end of the experiment, the temperature of the water is 21.5 °C.(a) The mass of the water in the beaker is 600 g and the specific heat capacity of water is4200 J / (kg °C).(i) Show that the increase in the internal energy of the water is 6300 J.[3](ii) In the experiment, the lamp is switched on for 500 s. The power supplied to the filamentlamp is 13 W. The useful energy from the lamp is transferred as light. The energy thatincreases the temperature of the water is wasted energy.Determine the maximum possible efficiency of the filament lamp.maximum possible efficiency = ......................................................... [4]90625/41/M/J/23©UCLES 2023 [Turn over(b) The efficiency of the lamp is less than the value determined in (a)(ii).Suggest one reason for this.................................................................................................................................................................................................................................................................................................................................................................................................................................................... [1]

Question

A student investigates the efficiency of a filament lamp. Fig. 4.1 shows the filament lamp with itsglass bulb immersed in water in a beaker.filament lampto power supplywaterbeakerthermometerFig. 4.1The 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 completelyinsulated.The lamp is switched on.At the end of the experiment, the temperature of the water is 21.5 °C.(a) The mass of the water in the beaker is 600 g and the specific heat capacity of water is4200 J / (kg °C).(i) Show that the increase in the internal energy of the water is 6300 J.3 In the experiment, the lamp is switched on for 500 s. The power supplied to the filamentlamp is 13 W. The useful energy from the lamp is transferred as light. The energy thatincreases the temperature of the water is wasted energy.Determine the maximum possible efficiency of the filament lamp.maximum possible efficiency = ......................................................... [4]90625/41/M/J/23©UCLES 2023 [Turn over(b) The efficiency of the lamp is less than the value determined in (a)(ii).Suggest one reason for this.................................................................................................................................................................................................................................................................................................................................................................................................................................................... [1]

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Solution

(a) (i) The increase in the internal energy of the water can be calculated using the formula:

Q = mcΔT

where: Q is the heat energy, m is the mass of the water, c is the specific heat capacity of the water, and ΔT is the change in temperature of the water.

Substituting the given values:

Q = (600 g / 1000 kg/g) * 4200 J/(kg°C) * (21.5°C - 19.0°C) Q = 0.6 kg * 4200 J/(kg°C) * 2.5°C Q = 6300 J

So, the increase in the internal energy of the water is 6300 J.

(a) (ii) The total energy supplied to the lamp can be calculated using the formula:

E = Pt

where: E is the energy, P is the power, and t is the time.

Substituting the given values:

E = 13 W * 500 s E = 6500 J

The efficiency of the lamp can be calculated using the formula:

Efficiency = (Useful energy / Total energy) * 100%

The useful energy is the energy that is not wasted, which is the total energy minus the wasted energy. In this case, the wasted energy is the energy that increased the temperature of the water, which is 6300 J.

So, the useful energy is:

Useful energy = Total energy - Wasted energy Useful energy = 6500 J - 6300 J Useful energy = 200 J

Substituting these values into the efficiency formula:

Efficiency = (200 J / 6500 J) * 100% Efficiency = 3.08%

So, the maximum possible efficiency of the filament lamp is 3.08%.

(b) The efficiency of the lamp is less than the value determined in (a)(ii) because not all the electrical energy supplied to the lamp is converted into light. Some of the energy is also converted into heat, which is not a useful form of energy in this case. This heat energy is lost to the surroundings, reducing the overall efficiency of the lamp.

This problem has been solved

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