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Digiron [165]
3 years ago
6

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 o

f 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?
Mathematics
1 answer:
FinnZ [79.3K]3 years ago
8 0

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  22minutes 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

       

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