How Do You Calculate the Efficiency of a 2.5kW Electric Kettle?

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SUMMARY

The efficiency of a 2.5kW electric kettle can be calculated by first determining the energy output of the kettle over a 3-minute period. Using the formula P = VI, where P is power (2.5 kW), the energy output is 450,000 J (2.5 kW * 180 seconds). The energy required to heat 1.2 kg of water from 20°C to 100°C is calculated using Q = mcΔT, resulting in 402 kJ. The efficiency is then calculated by comparing the energy used to heat the water to the energy output of the kettle, yielding an efficiency of approximately 89.3%.

PREREQUISITES
  • Understanding of basic thermodynamics concepts
  • Familiarity with the specific heat capacity of water
  • Knowledge of electrical power calculations (P = VI)
  • Ability to perform unit conversions (kW to J)
NEXT STEPS
  • Learn about thermodynamic efficiency calculations
  • Study the specific heat capacity of various substances
  • Explore electrical power formulas in greater detail
  • Investigate energy conservation principles in heating systems
USEFUL FOR

Students studying physics or engineering, electrical engineers, and anyone interested in the efficiency of heating appliances.

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Homework Statement


A 2.5kW electric kettle takes 3 mins to raise the temperature of 1.2kg of water from a temperature of 20 degrees C to a temperature of 100 degress. The specific heat capacity of water is 4.2kJ/(kg K)


Homework Equations


So far I know that we're either using P=vit ot Q=mct
How do i find:
(a) enerygy output from 2.5kW heating element in a time of 3 mins, (b) efficiency of the heating.
 
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A kw is 1000 J/second

Calculate how much energy it took to heat the water and compare that with the answer from A
 

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