differential equation?

Question

differential equations problem..


at 2:00 pm a thermometer reading 80 deg F is taken outside where the air temperature is 20 deg F. At 2:03 pm, the temperature reading yielded by the thermometer is 42 deg F.. Later the thermometer is brought inside where the air temperature 80 deg F.. At 2:10 pm, the temperature reading is 71 deg F. When was the thermometer brought indoors?

Answer 1

42F is (1-1/e) or 1 thermal time constant of the 80F-20F gap. The thermal time constant of the system, hence, is 3.0 minutes.

From 2:00 to 2:10 is 3.33 thermal time constants. Laying this 3.33 tau out: the downward path should take 1.65 time constants to get to 31.5F, at which point the thermometer should be taken indoors, and then after 1.68 time constants later its temperature will raise back from 31.5F to 71.0F. Counting the time backwards, 1.68 * 3.0, is 5.04 minutes.

Thus the thermometer was taken back indoors at (2:04:58.5 pm) about 2:05 pm.

Answer 2

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