# Question 32d48

Mar 21, 2015

To go from Joules to kWh you just have to use a simple conversion factor

3.86 * 10^(10)cancel("J") * (2.77778 * 10^(-7)"kWh")/(1cancel("Joule")) = 1.0722 * 10^(4)"kWh"#

Rounded to three sig figs, the answer will be $1.07 \cdot {10}^{4} \text{kWh}$.

Mar 21, 2015

I believe it's appropriate to explain the reasons behind the calculations presented in the first answer.

Let's start from definitions.

Joule (abbreviated as $J$) is a unit of energy equal to work made by a force of 1 newton ($1 N$) at a distance of 1 meter ($1 m$):
$1 J = 1 N \cdot 1 m$

Watt (abbreviated as $W$) is a unit of power (that is, energy per unit of time) equal to work of $1$ Joule ($1 J$) per second ($1 \sec$), that is $1 W = \frac{1 J}{1 \sec}$.

Therefore,
$1 W \cdot 1 \sec = 1 J$
or, in words, one Joule equals to one Watt-second.

Now we have to transform watt and second into kilowatt ($1 k W = 1000 W$) and hour ($1 h = 3600 \sec$):
$1 k W h = 1000 W \cdot 3600 \sec = 3600000 J$

Or, reversing this equality,
$1 J = \frac{1}{3600000} k W h = 2.777778 \cdot {10}^{-} 7 k W h$

Using the above equality, we do the calculations presented in the first answer:
$3.86 \cdot {10}^{10} J = \frac{3.86 \cdot {10}^{10}}{3600000} k W h =$
$= \frac{3.86 \cdot {10}^{10}}{3.6 \cdot {10}^{6}} k W h = 1.0722 \cdot {10}^{4} k W h$

As you see, all the difficulties of this problem lie in the definitions of the units of measurement: the two main ones ($1 J = 1 N \cdot 1 m$ - unit of energy and $1 W = \frac{1 J}{1 \sec}$ - unit of power) and a derived unit of kilowatt-hour, which is a power of $1000$ watt applied during the time of $1$ hour.