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Tasya [4]
3 years ago
12

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?
Mathematics
2 answers:
alex41 [277]3 years ago
4 0
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 <span>
38=9<span>e<span>^−.3344t</span></span></span> for t
this time we get

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

so 4.3 minutes before 2:10
ratelena [41]3 years ago
3 0

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

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