why is the efficiency of a calorimeter less than 100%
That would be shown as .01%. So you would multiply the number by .0001 to get one hundredth of a percent. One Hundredth of a percent of 1239 would be .1239, one tenth would be 1.239, one percent would be 12.39 and ten percent would be 123.9.
Well...if you drop a 0 for ten percent that would be 150. So half of that would be 5 percent which is 75.
Five Percent of 20 would be 1.
78/100
2 percent of 1000 would be 20.
In saying what the overall efficiency would be, I suppose you mean for other processes, creating the chemical energy for example, and using the thermal energy. This is impossible to answer, not knowing what these processes are.
A machine with 100 percent mechanical efficiency would be called an ideal machine, as it would have no energy losses due to friction, heat, or other inefficiencies.
Styrofoam Cup
Styrofoam Cup
Impurities in the substance can cause a greater percent yield. I recommend redoing the lab for better results.
You would burn it in a calorimeter :-)
The percent efficiency would be calculated by dividing the useful output by the total input energy and multiplying by 100. In this case, the useful output is 6W (light energy produced) and the total input is 60W (electricity consumed). Therefore, the efficiency would be (6/60) * 100 = 10%.
Nothing can be more that 100% (one hundred percent)! When people say that something is one hundred and ten percent - they are wrong. It is an attempt to imply that something is greater that it should be. It is similar to the annoying habit of adding !!!! at the end of a sentence, when a single ! is correct.
For a Carnot engine to achieve 100 percent efficiency, the temperature of the cold reservoir would need to be absolute zero (0 Kelvin). This is because the efficiency of a Carnot engine is given by 1 - (Tc/Th), where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir, and efficiency is maximized as Tc approaches absolute zero.
No heat loss = maximum output. There would be no loss of energy, which is an ideal condition.
72 percent
A ideal machine would have an efficiency of 100 percent. For this to be possible, the amount of energy output by the machine would equal the amount of energy input. Because all machines have physical parts, some energy is lost to friction, heat dissipation, or other factors, so no machine can be an ideal machine.