Question

It costs 160 cent for 3600 J of electrical energy. Determine how much it cost (in Rand) to boil the 750 ml of water in the kettle.

Answer 1

$3600J=0,001kV*h$ . It costs $160$ cent.

Energy of boiling of water:

$Q=rm=r\rho V$

$r=2.5*10^6\frac{J}{kg}$

$\rho=10^3{kg}{m^3}$

$V=750*10^{-6}m^3$

$Q=1875J=0.52kV*h$

Then

$0.001-160 0.52-x$

Taking the proportional and :

$x=\frac{0.52}{0.001}*160=83200$ cents.