Question

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

Answer 1

`t_1 = 2.8 min`

Explanation:

we know that

`T = T_s + (T_o -T_s)e^(-kt)`

From information given

At 2:00 PM

`T = 20 + (80 -20) e^(-kt)`

`T = 20 + 60e^(-kt)`

At 2:03 PM

`42 = 20 = 60e^(-k3)`

`e^(-k) = [(11)/(30)]^(1/3)`

hence

`T = 20 + 60 [(11)/(30)]^(t/3)`

`t_1 min` after 2:03. Thermometer reads

`T = 20 + 60[(11)/(30)]^(t_1/3)`

when instrument brought back to rom

`T = 80 + (T_o- 80)[(11)/(30)]^(t/3)`

`T = 80 + (20+ 60[(11)/(30)]^(t_1/3)- 80)[(11)/(30)]^(t/3)`

`T = 80+60[[(11)/(30)]^(t_1/3) -1] (11)/(30)^(t/3)`

AT 2.10 pm
`t = 7 - t_1`

`71 = 80+ 60[[(11)/(30)]^(t_1/3) -1] (11)/(30)^((7- t_1) /3)`

solving for t_1 we get

`t_1 = 2.8 min`

Answer 2

we want to know when we brought it back in to the room. at that time the temperature difference is 80 - 42 = 38 so we have to solve
38=9e^−.3344t for t
this time we get

38 / 9=e^−.3344t
t=ln(38 / 9) / −.3344 t=−4.3

so 4.3 minutes before 2:10