# Application of Differential Equation: Newton's Law of Cooling

Submitted by Mhar Jhune on March 3, 2017 - 2:33pm

At 2:00PM, a thermometer reading 80°F is taken outside where the air is 20°F. At 2:03PM, the temperature reading yielded by thermometer is 42°F, later the thermometer was brought inside where the air is at 80°F, at 2:10PM the reading is 71°F. When was the thermometer brought indoor?

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## $T = T_s + (T_o - T_s)e^{-kt}

$T = T_s + (T_o - T_s)e^{-kt}$

At 2:00PM

$T = 20 + (80 - 20)e^{-kt}$

$T = 20 + 60e^{-kt}$

At 2:03PM

$42 = 20 + 60e^{-k(3)}$

$e^{-3k} = \frac{11}{30}$

$e^{-k} = \left( \frac{11}{30} \right)^{1/3}$

Hence,

$T = 20 + 60\left( \frac{11}{30} \right)^{t/3}$

t

_{1}minutes after 2:03, the thermometer reads$T = 20 + 60\left( \frac{11}{30} \right)^{{t_1}/3}$

When the thermometer is brought back to the room

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

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

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

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

At 2:10PM, t = 7 - t

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

$-9 = 60 \left[ \left( \frac{11}{30} \right)^{t_1/3} - 1 \right] \dfrac{\left( \frac{11}{30} \right)^{7/3}}{\left( \frac{11}{30} \right)^{t_1/3}}$

$-9 = 60 \left[ \left( \frac{11}{30} \right)^{7/3} - \dfrac{\left( \frac{11}{30} \right)^{7/3}}{\left( \frac{11}{30} \right)^{t_1/3}} \right]$

$\dfrac{\left( \frac{11}{30} \right)^{7/3}}{\left( \frac{11}{30} \right)^{t_1/3}} = \left( \frac{11}{30} \right)^{7/3} + \dfrac{9}{60}$

$\dfrac{\left( \frac{11}{30} \right)^{7/3}}{\left( \frac{11}{30} \right)^{7/3} + \dfrac{9}{60}} = \left( \frac{11}{30} \right)^{t_1/3}$

$\left[ \dfrac{\left( \frac{11}{30} \right)^{7/3}}{\left( \frac{11}{30} \right)^{7/3} + \dfrac{9}{60}} \right]^3 = \left( \frac{11}{30} \right)^{t_1}$

$\ln \left[ \dfrac{\left( \frac{11}{30} \right)^{7/3}}{\left( \frac{11}{30} \right)^{7/3} + \dfrac{9}{60}} \right]^3 = \ln \left( \frac{11}{30} \right)^{t_1}$

$\ln \left[ \dfrac{\left( \frac{11}{30} \right)^{7/3}}{\left( \frac{11}{30} \right)^{7/3} + \dfrac{9}{60}} \right]^3 = t_1 \ln \left( \frac{11}{30} \right)$

$t_1 = 2.8 ~ \text{min}$

time = 2:05:48.6PM

## Where do you get the time= 2

Where do you get the time= 2:05:48.6 PM sir?

## First, convert 2.8 mins in

First, convert 2.8 mins in terms of minutes and second which would be equal to 2 minutes and 48 seconds (convert 0.8 min into seconds). Since his basis time is at 2:03 PM, you need to add the converted time to this time making it 2:05:46 PM.

## If ur wondering how he got

If ur wondering how he got the answer, try putting this in your calcu

2°03° + 0°2.8° = 2°5'48'' or 2:5.48

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