Question

A cup of tea with initial temperature of I = 85C^(circ) is left in the room with a constant temperature A = 21^(circ). In one minute the tea cools down to 84C^(circ). What will be the temperature of the tea after an hour? When will the temperature of the tea be 22^(circ)?

Answer 1

• 45.9° after 1 hour
• 22° after 264 minutes

Explanation:

The temperature can be modeled by a decaying exponential function with an initial value of (85° -21°) = 64° and a vertical offset of 21°. The temperature T as a function of time t can be written as ...

T(t) = 21 +64(63/64)^t . . . . where t is in minutes and T is in degrees

The ratio 63/64 is the fraction of the initial temperature difference that remains after 1 minute.

We are asked for two points on the curve described by this function:

• T(60)
• t when 22 = T(t)

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(a) T(60) = 21 +64(63/64)^60 ≈ 45.88

The temperature after 1 hour will be about 45.9 °C.

__

(b) Solving for t, we get ...

22 = 21 +64(63/64)^t . . . . fill in the value of T

1 = 64(63/64)^t . . . . . . . . . subtract 21

1/64 = (63/64)^t . . . . . . . . .divide by 64

log(1/64) = t·log(63/64) . . . take the log

log(1/64)/log(63/64) = t ≈ 264.08 . . . . divide by the coefficient of t

The temperature of the tea will be 22° after 264 minutes.