The temperature of a pan of hot water varies according to newton's law of cooling: dt dt equals negative k times the quantity t minus a, where t is the water temperature, a is the room temperature, and k is a positive constant.
if the water cools from 90°c to 85°c in 1 minute at a room temperature of 30°c, how long (to the nearest minute) will it take the water to cool to 60°c?

3

4

5

8

  8 minutes

Step-by-step explanation:

You can make a reasonable guess at the answer just using logic.

If the water cools at the constant rate of 5° per minute, it will take (90-60)/5 = 6 minutes for the water to cool to 60 °C. Since the rate of cooling decreases as the temperature decreases, it will take longer than 6 minutes to reach 60 °C.

The only answer choice that is longer than 6 minutes is 8 minutes.

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Newton's law of cooling results in an exponential function for the temperature. The initial temperature difference of 90-30 = 60 degrees decays to zero. We are told that 55/60 = 11/12 of that temperature difference remains after 1 minute, so the exponential equation can be written as ...

  T(t) = 30 +60·(11/12)^t . . . . . T in °C and t in minutes

This can be solved to find the value of t for T=60.

  60 = 30 +60(11/12)^t

  30/60 = (11/12)^t . . . . . . subtract 30, divide by 60

  log(1/2) = t·log(11/12) . . . take the log

  log(1/2)/log(11/12) = t ≈ 7.96617 ≈ 8 . . . . minutes

452,489 to the nearest thousand is 452,000

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