Calculating Energy and Efficiency for Heating Water in a Kettle

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To heat 1.5 L of water from 20 °C to 100 °C, 504 kJ of energy is required, calculated using the formula Q=m*Cp*T. When using a 2.0 kW cooking plate, it takes 7 minutes to reach boiling, resulting in an electrical energy consumption of 840 kJ. The efficiency of the heating process can be determined by comparing the useful energy output (504 kJ) to the total energy input (840 kJ), leading to a formula of efficiency = (output/input) x 100. The discussion highlights confusion regarding the definition of efficiency and the distinction between energy output and input. Ultimately, understanding the energy dynamics in heating water is crucial for calculating efficiency accurately.
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Homework Statement


The density of water is 1000 kg/m3 and the specific heat capacity of water is 4.2Kj/(Kg.C)

Homework Equations


a) How much energy is required to heat 1.5 L of water at a temperature of 20 °C to the boiling point 100 ºC?
b) There is 1.5 L of water in a kettle on a cooking plate of 2.0 kW. It takes 7.0 minutes to
heat the water from a temperature of 20 ºC to the boiling point. How much electrical
energy is required? What is the efficiency of the heating process?

The Attempt at a Solution


a)
Q=m*Cp*T
Q=1.5*4.2*(100-20)
Q=504 Kj.

b)
how can we find the electrical energy?
 
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chawki said:
how can we find the electrical energy?

...a cooking plate of 2.0 kW. It takes 7.0 minutes...

Do the obvious!
 
wow..is it like in cinematic..Power=work/time ?
so here we would have Power=energy/time?
and then 2000=E/(7*60)
E=840000 J = 840 KJ ?
 
Yes. Aren't general concepts wonderful?:biggrin:
 
Yes, amazing o:)
but what about part c, what efficiency they are talking about ?
i know it will probably be Q/E or E/Q...i'm not sure :rolleyes:
 
chawki said:
Yes, amazing o:)
but what about part c, what efficiency they are talking about ?
i know it will probably be Q/E or E/Q...i'm not sure :rolleyes:

Efficiency is the ratio of the (results/effort)x100. , or (output/input)x100, however you wish to look at it.

The water takes a certain amount of energy to boil as dictated by Nature. But in order to accomplish that result, the heater produced a much larger amount of energy. Obviously there was wasted energy and only part of the "efforts" of the heater went into accomplishing the desired result.
 
can you explain it more simply o:)
i just got that it's the enrgy output/input
but in our case, they said ''the heating process'' are they talking about the question a) ?
and where is the energy generated here :eek: which is supposed to be the Output ..i guess!
 
chawki said:
can you explain it more simply o:)
i just got that it's the enrgy output/input
but in our case, they said ''the heating process'' are they talking about the question a) ?
and where is the energy generated here :eek: which is supposed to be the Output ..i guess!

The output is hot water. How much energy went into the hot water?
The input is the energy provided by the heater. How much energy did it produce?
 
gneill said:
The output is hot water. How much energy went into the hot water?
The input is the energy provided by the heater. How much energy did it produce?

The energy that went into hot water is 504kj
but the question about how much energy did it produce..i'm not sure about it..probably that 840kj...but i don't see that is an energy produced
i think it's still an energy as the one of 504kj
 
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