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Newton’s law of cooling states that for a cooling substance with initial temperature t0, the temperature t(t) after t minutes can be modeled by the equation t(t)=ts+(t0−ts)e−kt, where ts is the surrounding temperature and k is the substance’s cooling rate. A liquid substance is heated to 80°c. Upon being removed from the heat, it cools to 60°c in 15 min. What is the substance’s cooling rate when the surrounding air temperature is 50°c? round the answer to four decimal places. 0. 0687 0. 0732 0. 0813 0. 872.

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The cooling rate of the substance is approximately 0.0732.

According to the statement, the Newton's law of cooling is defined by the following formula:

[tex]T(t) = T_{s} + (T_{o}-T_{s})\cdot e^{-k\cdot t}[/tex] (1)

Where:

  • [tex]T_{s}[/tex] - Final temperature, in degrees Celsius.
  • [tex]T_{o}[/tex] - Initial temperature, in degrees Celsius.
  • [tex]t[/tex] - Time, in minutes.
  • [tex]k[/tex] - Cooling rate, in [tex]\frac{1}{min}[/tex].
  • [tex]T(t)[/tex] - Current temperature, in degrees Celsius.

Please notice that substance reaches thermal equilibrium when [tex]T(t) = T_{s}[/tex], that is when temperature of the substance is equal to the temperature of surrounding air.

If we know that [tex]T_{o} = 80\,^{\circ}C[/tex], [tex]t = 15\,min[/tex], [tex]T_{s} = 50\,^{\circ}C[/tex] and [tex]T(15) = 60\,^{\circ}C[/tex], then the cooling rate of the substance is:

[tex]60 = 50 + (80 - 50)\cdot e^{-15\cdot k}[/tex]

[tex]\frac{60-50}{80-50}= e^{-15\cdot k}[/tex]

[tex]\frac{1}{3} = e^{-15\cdot k}[/tex]

[tex]k = -\frac{1}{15}\cdot \ln \frac{1}{3}[/tex]

[tex]k \approx 0.0732[/tex]

The cooling rate of the substance is approximately 0.0732.

To learn more on Newton's law of cooling, we kindly invite to check this verified question: https://brainly.com/question/13748261