I then plotted Ln(mass) against time in hours working on the assumption that the equation would be something like M=M0exp(-lambda.t). It's a much better straight line than I expected. I had thought that because break and lunchtime would be heavy on consumption, the graph would be distorted.
The gradient is -0.2332 /hr. The -gradient is the decay constant lambda. Now half life = Ln2/lambda = Ln2/0.2332 = 3 hours to 1 sf. Look at the graph below. It's not a bad fit.