Question

If a cup of coffee has temperature 97°C in a room where the ambient air temperature is 25°C, then, according to Newton's Law of Cooling, the temperature of the coffee after t minutes is T ( t ) = 25 + 72 e − t / 45 . What is the average temperature of the coffee during the first 22 minutes?

Answer 1

Answer:

The  average temperature  is $T_{a} = 81.95^oC$

Step-by-step explanation:

From the question we are told that

    The temperature of the coffee after time t is   $T(t) = 25 + 72 e^{[-\frac{t}{45} ]}$

Now the average temperature during the first 22 minutes i.e fro $0 \to 22$minutes is mathematically evaluated as

              $T_{a} = \frac{1}{22-0} \int\limits^{22}_{0} {25 +72 e^{[-\frac{t}{45} ]}} \, dx$

               $T_{a} = \frac{1}{22} [25 t + 72 [\frac{e^{[-\frac{t}{45} ]}}{-\frac{1}{45} } ] ] \left| 22} \atop {0}} \right.$

             $T_{a} = \frac{1}{22} [25 t - 3240e^{[-\frac{t}{45} ]} ] \left | 45} \atop {{0}} \right.$

              $T_{a} = \frac{1}{22} [25 (22) - 3240e^{[-\frac{22}{45} ]} - (- 3240e^{0} )]$

            $T_{a} = \frac{1}{22} [550 - 1987.12 + 3240]$

          $T_{a} = 81.95^oC$