Put your computers on a UPS, even a small one. Just protecting yourself from short power outages will save you a bunch of downtime. Even if you resolve your overload issue, this is still worth it.
Most computer users today can get by with a laptop that's under $500, giving you built-in battery backup & portability in a compact, low-power package. Plug it in to your keyboard, mouse, and monitor the same as your desktop today, so your work experience doesn't change.
The power strip you linked to (http://www.cyberpowersystems.com/products/surge-protectors/home-surge/6050S.html?selectedTabId=specifications&imageI=#tab-box) doesn't appear to have a breaker. I think that whoever wrote that was just confused about what it means to be rated for 15A.
If your coffee maker and space heater are in the same location, you could plug them into mutually exclusive switched outlets. You'll need:
A steel square box, and an appropriate face plate
A regular duplex receptacle. 15A or 20A can work
a 3-way switch
cable
plug (15A or 20A, to match the recep)
fittings
Snap off the tab on the hot side of the recep, then run short leads from those 2 screws to the 3-way switch. This will let the switch choose one socket or the other. Plug in the heater on one and the coffee maker on the other. Now it's easy to make sure only one is in use at a time.
(Later I will add some pictures and other details. If anyone has pointers to the correct fittings and cable, please comment.)
Power adapters convert the electricity from one voltage to another voltage, and sometimes between AC and DC.
The quick answer to your question is to multiply the INPUT voltage by the INPUT current. In this case, you likely have 120 V input (if you're in the United States). So, the MAXIMUM power would be 120*0.24 = 29 W, though the actual power usage is likely less than this. With efficient power supplies, it will be better to look at the electrical requirement of your device rather than the power supply's maximum power rating. For example, if the device uses 1.2 A at 5 V DC, it would use a maximum of about 6 W. I would multiply this number by about 1.2 to account for power supply inefficiencies, so your system may use about 7.2 W max.
And here's some related information about power supply efficiency that is somewhat relevant to the question (that I wrote before carefully reading the question):
The INPUT is what sort of electrical system you need to supply to the adapter (i.e. what your power company supplies).
The OUTPUT is what is supplied to your device.
Note that he amount of DC electrical power is calculated by multiplying the current by the voltage (P=I·V). For AC, this product is the maximum power that could be used, though the actual amount could be lower because of the power factor (P=I·V·PF), and the power factor is typically between 0.7 and 1.0 (except for some motors).
Because of inefficiencies in the power adapter not all of the input power is able to be output. This extra power is turned into heat. So, the input power is always larger than the output power.
In your example, the power adapter is rated to use a maximum of 58 W (0.24*240), but can output only 5*1.5=7.5 W. So, worse case based on the label, it will use 58 W, but only supply 7.5 W to your device, so about 13% efficient. Efficiency is defined as the output power divided by the input power.
Power supplies usually do not draw their maximum power during usage: They try to draw only the amount of power the device wants. So, this adapter usually won't be drawing 58 W. In fact, modern power supplies will use less than 1 W when plugged in with their device turned off and have inefficiencies of >90%. So, knowing the maximum does not tell you their typical usage.
One quick test of the efficiency is to check the temperature of the power supply. The hotter the power supply (while plugged in), the less efficient it is.
When choosing a power supply to reduce energy use, pay attention to their efficiency and not their energy use. New supplies should be rated as to their energy usage using roman numerals (I,II,III,IV,V) based on an international standard. Class V is currently the best efficiency rating, and will mean that the drawn power will closest to the supplied power, in which case you should look at the devices required voltage and current to get an idea of the power used.
Best Answer
I looked at this again now that it has tripped once more with 38 degrees F outside temperature. It is an AFCI breaker, not a GFCI (or possibly more correct it is both), and when flipping it back to on, the light labeled ARC FAULT comes on (not the other one labeled GROUND FAULT).
Upon more reading, simply ensuring the microwave isn't underloaded (e.g. add a cup of water to it while heating something small) should prevent this.
Except that what is funny about this is that when it tripped today it in fact already had a bit of extra water inside. Although it was a shot glass of water, not a whole cup. Now that I'm barking up the right tree I should be able to get somewhere with this though.