Heating of a liquid

A liquid with dissolved solids was placed on a stove at a certain setting and the temperature of the liquid was measured over time. The data was graphed (temperature vs time) which gave a curved line that maxed out. A best fit of the data yielded an equation for the line, and the first derivative of that equation gave a rate of change of the temperature per unit time at each time point:

f '(t)=30e^−0.3t

This equation describes how the liquid responds to the heat setting of the stove.

For this experiment, the amount that the temperature increased from 0 to 5 minutes can be calculated using integrals, which can give total amounts:

∫f '(t)dt = ∫30e^−0.3tdt

= (-30/0.3)(e^−0.3*5 - e^0.3*0) this result is found by looking up the integral ∫ae^−btdt in a table of integrals (or here) and using the boundary conditions 0 and 5.

= 77.7 degrees

in conclusion, integrating a rate of change equation will give the total change!