I have created a simple 50 Ω dummy load to test transmitters. I also added a simple RF diode detector so I can measure the peak voltage, and calculate the power.
The dummy load consists of eight 100 Ω resistors rated at 2 W so the load should handle 16 W, at least for short periods. I constructed the dummy load using a combination of ugly construction and Manhattan style, by gluing pieces of PCB (as isolation pads) on top of a ground plane PCB. Then I soldered the components directly on the copper without drilling holes.
The RF probe part consist of a simple 1N4148 diode and a 0.01 uF ceramic capacitor. I only had a 50 V capacitor in my junk box, but it should be sufficient given that this is a 16 W dummy load and 16 W translates to 40 V peak.
I added female banana connectors, which connects to a multimeter. The power can be calculated by Ohms law by subtracting the forward voltage on the diode from the measured voltage, then multiply by 0.707 (to get RMS), then square the result and finally divide by 50 Ω. Some homebrewers add a voltage divider to their RF probes using a 4M7 resistor to get RMS voltage directly. I did not bother since I am sure the input impedances on my multimeters varies.
The calculations might seem a bit cumbersome, but I might print out a small lookup table and glue it to the box to have some ballpark figures. However, this is not a precision instrument. The forward voltage on the diode varies with load and can be somewhere between 0.4 and 0.7 V. I simply use 0.5 V in my measurements.