Difference between revisions of "Heating of a liquid"
| Line 11: | Line 11: | ||
= -30/0.3(e<sup>−0.3*5</sup>-e<sup>0</sup>) | = -30/0.3(e<sup>−0.3*5</sup>-e<sup>0</sup>) | ||
| − | this result is found by looking up the integral <font size = "+2"><span>∫</span></font>ae<sup>−bt</sup>dt and using the boundary conditions 0 and 5. | + | this result is found by looking up the integral <font size = "+2"><span>∫</span></font>ae<sup>−bt</sup>dt in a table of integrals and using the boundary conditions 0 and 5. |
Revision as of 16:44, 1 April 2021
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:
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-e0)
this result is found by looking up the integral ∫ae−btdt in a table of integrals and using the boundary conditions 0 and 5.
