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.
78/100
Five Percent of 20 would be 1.
2 percent of 1000 would be 20.
If the efficiency of converting chemical energy to thermal energy is 90 percent, the overall efficiency would depend on the efficiency of converting thermal energy to the desired output (e.g., mechanical energy or electricity). Generally, the overall efficiency would be lower than 90 percent due to losses in the subsequent conversion processes.
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.
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%.
No, a machine cannot have 110 percent efficiency. Efficiency is calculated as output divided by input, with 100 percent being the maximum achievable efficiency. Having a value higher than 100 percent would imply that the machine is producing more output than the input provided, which is not physically possible.
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.
You would burn it in a calorimeter :-)
72 percent
No heat loss = maximum output. There would be no loss of energy, which is an ideal condition.
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.