Calculating Energy and Efficiency for Heating Water in a Kettle

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In summary, the specific heat capacity of water is 4.2Kj/(Kg.C) and the density of water is 1000 kg/m3. The first question asks how much energy is required to heat 1.5 L of water from 20 °C to 100 °C, which is calculated to be 504 Kj. The second question asks for the electrical energy required to heat the water using a 2.0 kW cooking plate for 7.0 minutes, which is calculated to be 840 Kj. The efficiency of the heating process is the ratio of output (hot water) to input (energy provided
  • #1
chawki
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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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  • #2
chawki said:
how can we find the electrical energy?

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

Do the obvious!
 
  • #3
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 ?
 
  • #4
Yes. Aren't general concepts wonderful?:biggrin:
 
  • #5
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 :uhh:
 
  • #6
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 :uhh:

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.
 
  • #7
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!
 
  • #8
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?
 
  • #9
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
 

1. How do I calculate the energy required to heat water in a kettle?

To calculate the energy required to heat water in a kettle, you will need to know the mass of the water (in kilograms), the specific heat capacity of water (4.186 J/g·°C), and the desired temperature change. You can use the formula Q = m x c x ΔT, where Q is the energy in joules, m is the mass of the water, c is the specific heat capacity, and ΔT is the temperature change in degrees Celsius.

2. How can I determine the efficiency of a kettle in heating water?

The efficiency of a kettle in heating water can be determined by dividing the energy output (the energy used to heat the water) by the energy input (the energy used by the kettle). Multiply the result by 100 to get a percentage. For example, if a kettle uses 1000 joules of energy to heat 500 grams of water and only 800 joules actually go into heating the water, the efficiency would be (800/1000) x 100 = 80%.

3. Does the material of the kettle affect the energy and efficiency of heating water?

Yes, the material of the kettle can have an impact on both the energy and efficiency of heating water. Materials with high thermal conductivity, such as copper or aluminum, will heat water more quickly and efficiently compared to materials with lower thermal conductivity, such as glass or ceramic.

4. Is it more energy efficient to heat a full kettle of water or just the amount needed?

It is generally more energy efficient to heat only the amount of water needed rather than heating a full kettle. This is because heating a smaller amount of water requires less energy and time compared to heating a larger amount. However, if you frequently use hot water, it may be more convenient to heat a full kettle and use the excess water for future use.

5. How can I improve the energy efficiency of heating water in a kettle?

To improve the energy efficiency of heating water in a kettle, you can try using a kettle with a higher thermal conductivity material, heating only the amount of water needed, and using a lid to prevent heat loss. Additionally, regularly descaling your kettle can also improve its efficiency by reducing buildup on the heating element.

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