00:01
In this problem, we'll start by calculating the efficiency of a professor outputting 2 .1 times 10 of the 5 joules with a 500 -kilocalorie energy input.
00:11
To do this, we just need to know the formula for efficiency is the output energy divided by the input energy, and we need to know the conversion factor between kilocalories and joules.
00:24
To do this, i've written that 4 .184 kilojoules is 1 kilocalory, and i will calculate this, first value, 2 .1 times 10 of the 5th joules.
00:36
I'll rewrite that as simply 2 .1 times 10 squared kilojoules.
00:42
To invert this, i've just reduced the power in the exponent by 3.
00:46
10 of the 5th goes down to 10 to the second power because joules to kilojoules is simply division by 1 ,000.
00:53
So this is the output.
00:57
And now we have to do 500 kilo calories into kilojoules.
01:07
We know one kilo calorie is equal to 4 .184 kilojoules.
01:12
So we'll have the one kilo calorie in the bottom because that's what we want to cancel out.
01:17
And the 4 .184 kilojoules in the numerator.
01:22
So that's what we'd like to be left with.
01:24
Okay, we just have to multiply these two values.
01:27
500 times 4 .184 gets us 2 ,092 kilojoules...