Question

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?

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Answer 1

8 minutes

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