Question

In an evaporation of water experiment, the initial mass of water in a beaker was 10.5775 +/- 0.0002 g. After some time, the mass of the water in the beaker was 10.3005 +/- 0.0002 g. The mass evaporated is calculated by subtracting the final mass from the initial mass. What is the final propagated error on the evaporated mass?

Answer 1

0.0003

Step-by-step explanation:

I created an account just to say that the other guy is wrong and cost me points on a quiz. I know I should have just done it myself from the start but I'm lazy and easily doubt myself. Anyway, you get the answer by taking the square root of (.0002)^2 + (.0002)^2, like so
`\sqrt{(.0002)^2+ (.0002)^2`. The answer is 0.0003.

Answer 2

Step-by-step explanation:

The theory of propagation of error in case of addition and subtraction states that maximum errors are added in absolute terms in both the operation of addition and subtraction . So in this case the subtracted value will be

10.5775 - 10.3005

= .2770 g

errors will be added ie in subtracted value we can find error to the tune of

.0002 + .0002 = .0004 g

So the subtracted value will be written as

.2770 ± .0004