1
$\begingroup$

Considering water of 20 litres 25 degree celsius of temperature needs to be raised. Using the heat equation q=mcpdt considering cp of water to be 4.2 kJ/kgk we get q=2100 kJ which equals 2100000 J. It means this amount of heat is required for heating the water under said conditions.

I considered 1500 W of heating coil made up of aluminium which equals 1500 J/s. So for this coil to generate 2100000 J of heat (2100000 J)/ (1500 J/s)=1400 secs =23.333 mins. Here to find the electric power consumption in kWh (1.5 kW)*(23.333/60)=0.583 kWh

Again if I consider 3000 W of heating coil made up of some other material say copper to generate the same amount of heat of 2100000 J the time required will be (2100000 J)/ (3000 J/s)= 700 secs= 11.66 mins. Electric power consumption in kWh (3 kW)*(11.66/60)= 0.583 kWh

The power consumption in both the coils of same dimension is same (0.583 kWh) as per calculations(only if I did it right). Does it specify that the neither the material of the coil nor the kW rating affect the power consumption to produce the said amount of heat ? Does it mean that the aluminium coil is as efficient as copper coil ? Where did the high and low resistances offered by the material for heating worked out in this calculation ? Does it mean resistance only affects the power rating and not the consumption ? Highly confused.

$\endgroup$
  • 1
    $\begingroup$ You are fixed on an unimportant detail. You heat the same amount of water, you need the same exact amount of energy (kWh). Nevermind the method or whatever... you see what you missed? $\endgroup$ – jaromrax Feb 19 '18 at 12:13
0
$\begingroup$

What you have done here is essentially an elaborate unit conversion. $\mathrm{kWh}$ is a unit of power ($\mathrm{kW}$) times a unit of time ($\mathrm{h}$), which makes in a unit of energy. In your calculation you have worked out the time needed for a certain amount of energy to be delivered by the element $$ t = \frac{E}{P} $$ and them multiplied that time by the power of the heating element, in slightly different units, getting you back to the total energy you put in (only you are now meassuring energy in $\mathrm{kWh}$ rather than $\mathrm{J}$). $$ E = tP $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.