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

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

Answer 2

Answer:

$t_1 = 2.8 min$

Step-by-step 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

$42 = 20 = 60e^{-k3}$

$e^{-k} = [\frac{11}{30}]^{1/3}$

hence

$T = 20 + 60 [\frac{11}{30}]^{t/3}$

$t_1 min$ after 2:03. Thermometer reads

$T = 20 + 60[\frac{11}{30}]^{t_1/3}$

when instrument brought back to rom

$T = 80 + (T_o- 80)[\frac{11}{30}]^{t/3}$

$T = 80 + (20+ 60[\frac{11}{30}]^{t_1/3}- 80)[\frac{11}{30}]^{t/3}$

$T = 80+60[[\frac{11}{30}]^{t_1/3} -1] \frac{11}{30}^{t/3}$

AT 2.10 pm $t = 7 - t_1$

$71 = 80+ 60[[\frac{11}{30}]^{t_1/3} -1] \frac{11}{30}^{(7- t_1) /3}$

solving for t_1 we get

$t_1 = 2.8 min$