Question

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.

Answer 1

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:

$T(t) = T_{s} + (T_{o}-T_{s})\cdot e^{-k\cdot t}$ (1)

Where:

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

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

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

$60 = 50 + (80 - 50)\cdot e^{-15\cdot k}$

$\frac{60-50}{80-50}= e^{-15\cdot k}$

$\frac{1}{3} = e^{-15\cdot k}$

$k = -\frac{1}{15}\cdot \ln \frac{1}{3}$

$k \approx 0.0732$

The cooling rate of the substance is approximately 0.0732.

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