# Required mass of wood to produce 20 Watts/Hour

Requirements:

• I need to produce $20\ \mathrm{W}\,\mathrm{hr}^{-1}$ by burning wood.
• The specific energy of wood is $21\ \mathrm{MJ}\,\mathrm{kg}$
• Find the required mass of wood to produce $20\ \mathrm{W}\,\mathrm{hr}^{-1}$
• Assume ideal conditions

What I tried:

$P=E/T$ where $P = 20\ \mathrm{W}\,\mathrm{hr}^{-1}$ and $T = 1\ \mathrm{hr} = 3600\ \mathrm{s}$, which yields $E=72\ \mathrm{kJ}$ when rearranged.

Now I know that wood has a specific energy of $21\ \mathrm{MJ}\,\mathrm{kg}^{-1}$, thus specific energy is $E/m$ where $E=72\ \mathrm{kJ}$ as found above, which yields $21\times 10^6 = 72\times10^3/m$.

Solving for $m$ gives 3.4 milligrams of wood. I appear to have missed something.

What have I missed?

• I'm confused by your units of $W\,hr^{-1}$. Do you mean watt-hours ($W \, hr$) as in energy, or just watts, as in power? – Carlton Oct 28 '15 at 1:14
• Seconding what @Carlton has said. You're using a nonsensical unit for power. Watts are already per unit time (1 W = 1 J/s). W/hr would I guess be a rate of change of power? It's certainly not a measure of power though, so that's your first problem. – Trevor Archibald Oct 28 '15 at 14:10
• I feel a bit silly I didn't notice that when editing. I guess my brain glossed over it and back-filled it with regular old Watts when I came up with my answer. – wwarriner Oct 29 '15 at 13:54

There it is! The left-hand-side of the specific energy equation is $21\times 10^6\ \mathrm{J}\:\mathrm{kg}^{-1}$, but the final units you've written are $\mathrm{mg}$. However, the $\mathrm{kg}$ would carry through since no conversion is noted, so that $m=3.4\times 10^{-3}\ \mathrm{kg}$, which is $3.4\ \mathrm{g}$, a more reasonable answer for perfectly efficient combustion of organic material.
As noted in the comments, one of the desired values is called power, but has units of $\mathrm{W}\,\mathrm{hr}^{-1}$, which is rate of change of power. My answer is predicated on the assumption the desired value is $20\ \mathrm{W}$.