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 mi
nute 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)?
1 answer:
Answer:
- 45.9° after 1 hour
- 22° after 264 minutes
Step-by-step 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.
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(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.
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