Differential Equation: Thermometer reading

Submitted by Ace Robert Campos on

At 1:00pm., a thermometer reading 70F is taken outside where the air temperature is -10F (ten below zero). At 1:02pm., the reading is 26F. At 1:05pm., the thermometer is taken back indoors, where the air is at 70F. What is the temperature reading at 1:09pm?

Tags
Heat Transfer
Thermometer Reading

Jhun Vert Thu, 03/03/2016 - 15:21

Thermometer was taken outside

$T = T_s + (T_o - T_s)e^{-kt}$   ←   derivation here.

$T = -10 + (70 + 10)e^{-kt}$

$T = -10 + 80e^{-kt}$
 

At 1:02 pm, t = 2 and T = 26°F
$26 = -10 + 80e^{-2k}$

$26 = -10 + 80e^{-2k}$

$\frac{36}{80} = e^{-2k}$

$e^{-k} = \left( \frac{9}{20} \right)^{1/2}$
 

Hence,
$T = -10 + 80\left( \frac{9}{20} \right)^{t/2}$
 

At 1:05pm, t = 5
$T = -10 + 80\left( \frac{9}{20} \right)^{5/2}$

$T = 0.8673^\circ F$   ←   thermometer reading at 1:05 pm

 

Thermometer was brought back to the room

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

$T = 70 + (0.8673 - 70)\left( \frac{9}{20} \right)^{t/2}$

$T = 70 + (0.8673 - 70)\left( \frac{9}{20} \right)^{t/2}$

$T = 70 - 69.1327\left( \frac{9}{20} \right)^{t/2}$
 

At 1:09 pm, t = 4
$T = 70 - 69.1327\left( \frac{9}{20} \right)^{4/2}$

$T = 56^\circ F$           answer