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.
78/100
Five Percent of 20 would be 1.
2 percent of 1000 would be 20.
an ideal machine
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.
Styrofoam Cup
Styrofoam Cup
Impurities in the substance can cause a greater percent yield. I recommend redoing the lab for better results.
72 percent
No heat loss = maximum output. There would be no loss of energy, which is an ideal condition.
You would burn it in a calorimeter :-)
Because unavoidably there would be a loss of energy during the process
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.
The specific heat of water is different from the specific heat of ice and so 'wet ice' into a calorimeter experiment can increase the mass of water in the calorimeter and become a source of unaccuracy.
If this, ie 70%, is the total energy loss, it follows that the engine is producing 30% useful energy, so that is its thermodynamic efficiency.