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

The average temperature is
`T_(a) = 81.95^oC`

Explanation:

From the question we are told that

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

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

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

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

`T_(a) = (1)/(22) [25 t - 3240e^{[-(t)/(45) ]} ] \left | 45} \atop {{0}} \right.`

`T_(a) = (1)/(22) [25 (22) - 3240e^{[-(22)/(45) ]} - (- 3240e^(0) )]`

`T_(a) = (1)/(22) [550 - 1987.12 + 3240]`

`T_(a) = 81.95^oC`