Do Microwave oven cook times grow linearly with Wattage? Calculating optimal cook time

So this is a completely random and trivial question that was prompted by looking at my microwave oven and the back of a TV dinner and my google searching failed to produce a meaningful answer so figured I'd ask here.

On my TV dinner box it has different cook times based on Microwave Oven Wattage:

1100 Watts - Cook 2 minutes, stir and cook for another 1.5 minutes. (3.5 minutes total)

700 Watts - Cook 3 minutes, stir and cook for another 2.5 minutes. (5.5 minutes total)

My oven is 900 Watts, which is right in the middle.

Assuming those times listed on the box are the scientifically optimal cook time (which is doubtful, but just go with me), is it fair to assume I should use the linear average for 2.5 minutes, stir and cook for another 2 minutes (4.5 minutes total), or is there a different rate of growth between the 700 watt and 1100 watt ovens that would change the optimal cook time?

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One way to test this is using a measured quantity of water (e.g 100mL) and see how long it takes it to boil when using different wattages. –  NWard Jun 4 '13 at 19:54

The rate at which a mass absorbs microwave radiation is characterized by the 'Specific Absorption Rate', which is proportional to the electromagnetic field intensity:

Wikipedia has a dedicated article to this phenomenon but in short it says

$$\text{SAR} = \int_\text{sample} \frac{\sigma(\vec{r})|\vec{E}(\vec{r})|^2}{\rho(\vec{r})} d\vec{r}$$

Because the absorption rate is proportional to the EM field intensity, $|\vec{E}(\vec{r})|^2$, which is in turn proportional to power, then the relationship will indeed be linear.

Assuming 100% energy efficiency (which is a wild overestimate: 20% might be more accurate but I do not know the answer to that question) your total 'energy' transferred to your dinner will be:

$$\text{Energy} = \text{Power} \Delta t$$

i.e.

$$\Delta t = \frac{\text{Energy}}{\text{Power}}$$

The cook time will be inversely proportional to your oven power.

1100 watts for 3.5 minutes computes to $$\text{Energy } = 1100 \text{ Watts } \times 210 \text{ seconds } = 231,000 \text{ Joules}$$

700 Watts for 5.5 minutes computes to $$\text{Energy } = 700 \text{ Watts } \times 330 \text{ seconds } = 231,000 \text{ Joules}$$

Thus a 900 Watt oven would necessitate

$$\Delta t = \frac{\text{Energy}}{\text{Power}} = 256.66 \text{ seconds } = 4.278 \text{ minutes }$$

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Thanks. That was clearly explained. –  David Stinemetze Jun 4 '13 at 21:56

To a first approximation, the product of power $P$ and time $t$ should be constant, since the energy put into heating the object is $$E = Pt.$$ That is, to achieve the same effect, time should be inversely proportional to power. (Of course, if the power is so low that the process lasts for hours, one would need to consider the rate at which the food is cooling as well.)

Note that $$(1100\ \mathrm{W}) \times (210\ \mathrm{s}) = 231\ \mathrm{kJ},$$ while $$(700\ \mathrm{W}) \times (330\ \mathrm{s}) = 231\ \mathrm{kJ},$$ so our assumption of constant energy seems to be what the manufacturer used. Then with a $900\ \mathrm{W}$ power source the required time is $$t = \frac{E}{P} = \frac{231\ \mathrm{kJ}}{900\ \mathrm{W}} = 257\ \mathrm{s}$$ total, to the nearest second. Note that this is not quite the mean of $210\ \mathrm{s}$ and $330\ \mathrm{s}$.

By analogy, think of this in terms of distances (rather than energies), speeds (rather than powers), and times. If a destination is $12$ miles away, you can walk at $1\ \mathrm{mph}$ for $12$ hours, $3\ \mathrm{mph}$ for $4$ hours, or $2\ \mathrm{mph}$ for $6$ (but not $(12+4)/2 = 8$) hours.

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