Evaluating a value for the density of water

 

I put my empty jug on the scales. The resolution of the digital scale is +- 0.5 grams because another half a gram would tip it over to 76 grams. 

I added 100ml of water. This is the same as 100 cubic centimetres. The resolution of the scale is one scale division and the bottom of the meniscus should be sat on the line. So I have 100+-25cm^3 which has a 25% uncertainty. Very high.

I put the jug on the scales again. To find the mass of the water, I have to subtract the mass of the jug. 164 - 75 = 89 grams. Now when I am ADDING OR SUBTRACTING amounts, I always ADD the +- uncertainty, which is called the ABSOLUTE UNCERTAINTY (not percentage uncertainty). It was 164+-0.5grams and 75+-0.5grams, so the result is a mass of water which is 89+-1gram. 

The percentage uncertainty of the mass is 1/89 x 100 = 1.1%

To calculate the density I do mass/volume = 89/100 = 0.89 grams per cubic centrimetre.

When I MULTIPY or DIVIDE, I always ADD %U.

%U for mass + %U for volume = 1.1 + 25 = 26% (2sf)

Now 26% of 0.89 = 0.23 so I can write that my value for the density of water is 0.89+-0.23g/cm^3.

The true value for the density of water is 1g/cm^3, which lies within the uncertainty of my measurement.

I repeated the experiment with a larger volume:

This time I got 0.98+-0.05 grams per cubic centimetre. This shows that by measuring larger amounts with the same equipment, we reduce the percentage uncertainties and thus the final absolute uncertainty.

YOU NEED TO SHOW HOW I WORKED OUT THIS LAST RESULT FROM THE READINGS IN THE PHOTOGRAPHS!