Difference between revisions of "Heating of a liquid"

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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:
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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<sup>−0.3t</sup>
 
f '(t)=30e<sup>−0.3t</sup>
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This equation describes how the liquid responds to the heat setting of the stove.
 
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.
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For this experiment, the amount that the temperature increased from 0 to 5 minutes can be calculated using integrals, which can give total amounts:
  
 
<font size = "+2"><span>&#8747;</span></font>f '(t)dt = <font size = "+2"><span>&#8747;</span></font>30e<sup>−0.3t</sup>dt
 
<font size = "+2"><span>&#8747;</span></font>f '(t)dt = <font size = "+2"><span>&#8747;</span></font>30e<sup>−0.3t</sup>dt
  
= -30/0.3(e<sup>−0.3*5</sup>-e<sup>0</sup>)
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= (-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>&#8747;</span></font>ae<sup>−bt</sup>dt in a table of integrals (or [https://www.integral-calculator.com/ here]) and using the boundary conditions 0 and 5.
  
this result is found by looking up the integral <font size = "+2"><span>&#8747;</span></font>ae<sup>−bt</sup>dt in a table of integrals and using the boundary conditions 0 and 5.
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= 77.7 degrees
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in conclusion, integrating a rate of change equation will give the total change!

Latest revision as of 17:14, 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 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 - e0.3*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!