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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?

Verified Answer :

syed514

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:

\(t_1 = 2.8 min\)

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

\(t_1 = 2.8 min\)