Difference between revisions of "Heating of a liquid"
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<font size = "+2"><span>∫</span></font>f '(t)dt = <font size = "+2"><span>∫</span></font>30e<sup>−0.3t</sup>dt | <font size = "+2"><span>∫</span></font>f '(t)dt = <font size = "+2"><span>∫</span></font>30e<sup>−0.3t</sup>dt | ||
| − | = -30/0.3(e<sup>−0.3*5</sup> - e<sup>0.3*0</sup>) 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. | + | = (-30/0.3)(e<sup>−0.3*5</sup> - e<sup>0.3*0</sup>) 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. |
= 77.7 degrees Celsius | = 77.7 degrees Celsius | ||
Revision as of 16:48, 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 - e0.3*0) this result is found by looking up the integral ∫ae−btdt in a table of integrals and using the boundary conditions 0 and 5.
= 77.7 degrees Celsius
