Difference between revisions of "Heating of a liquid"
Jump to navigation
Jump to search
| Line 1: | Line 1: | ||
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: | 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<sup>−0.3t</sup> | + | f'(t)=30e<sup>−0.3t</sup> |
This equation describes how the liquid responds to the heat setting of the stove. In this case, f(t) is the temperature change per minute and t is the time in minutes. | This equation describes how the liquid responds to the heat setting of the stove. In this case, f(t) is the temperature change per minute and t is the time in minutes. | ||
Revision as of 16:37, 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. In this case, f(t) is the temperature change per minute and t is the time in minutes.
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
