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Answer :
To determine the energy efficiency of the kettle, we need to calculate the amount of energy it transferred to the water and compare it to the electrical energy it consumed. The power rating of the kettle is given as 900 W, which means it consumes 900 Joules of electrical energy per second.
First, we need to convert the time into seconds. 4 minutes and 30 seconds is equivalent to 270 seconds.
Next, we calculate the change in temperature of the water. The initial temperature is 24°C, and it increases to 48°C. So the temperature change is 48°C - 24°C = 24°C.
The energy transferred to the water can be calculated using the formula:
Energy = mass of water × specific heat capacity of water × temperature change
The mass of water is given as 1.0 L, which is equivalent to 1.0 kg since the density of water is approximately 1 g/cm³. The specific heat capacity of water is approximately 4.18 J/g°C.
Energy transferred to water = 1.0 kg × 4.18 J/g°C × 24°C = 100.32 kJ
Finally, we can determine the energy efficiency by dividing the energy transferred to the water by the electrical energy consumed:
Energy efficiency = (Energy transferred to water / Electrical energy consumed) × 100
= (100.32 kJ / (900 W × 270 s)) × 100
≈ 0.412 or 41.2%
Therefore, the kettle has an energy efficiency of approximately 41.2%
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