# Differential equations: Newton's Law of Cooling

Submitted by John Philip on September 20, 2019 - 4:46pm

Need help on this one. DE Newton's Law of Cooling

At 9 A.M., a thermometer reading 70°F is taken outdoors where the temperature is

l 5°F. At 9: 05 A.M., the thermometer reading is 45°F. At 9: 10 A.M., the thermometer is

taken back indoors where the temperature is fixed at 70°F. Find (a) the reading at

9: 20 A.M. and (b) when the reading, to the nearest degree, will show the correct

(70°F) indoor temperature.

## New forum topics

- Please help me solve this problem: Moment capacity of a rectangular timber beam
- Solid Mensuration: Prismatoid
- Differential Equation: (1-xy)^-2 dx + [y^2 + x^2 (1-xy)^-2] dy = 0
- Differential Equation: y' = x^3 - 2xy, where y(1)=1 and y' = 2(2x-y) that passes through (0,1)
- Tapered Beam
- Vickers hardness: Distance between indentations
- Time rates: Question for Problem #12
- Make the curve y=ax³+bx²+cx+d have a critical point at (0,-2) and also be a tangent to the line 3x+y+3=0 at (-1,0).
- Minima maxima: Arbitrary constants for a cubic
- Minima Maxima: y=ax³+bx²+cx+d

## Newton's law of Cooling:

Newton's law of Cooling:

$\dfrac{dT}{dt} = k(T - T_s)$

$\dfrac{dT}{T - T_s} = k \, dt$

$\ln (T - T_s) = kt + \ln C$

$\ln (T - T_s) = \ln e^{kt} + \ln C$

$\ln (T - T_s) = \ln Ce^{kt}$

Hence,

$T - T_s = Ce^{kt}$

At 9:00 am (

t= 0 ), the thermometer was taken outdoors whereT=15°F and the thermometer reading was_{s}T= 70°F.$70 - 15 = Ce^{k(0)}$

$C = 55$

Therefore,

$T - 15 = 55e^{kt}$

At 9:05 am (

t= 5), the thermometer reads 45°F:$45 - 15 = 55e^{k(5)}$

$k = -0.121\,227\,160\,7$

At 9:10 am (

t= 10), the thermometer is taken back indoors.$T - 15 = 55e^{-0.121\,227\,160\,7(10)}$

$T = 31.364^\circ F$ ← thermometer reading just before it was taken back indoors

Inside the room, the surrounding temperature

Tis 70°F, and we will reset the_{s}tto zero, hence,$T - T_s = Ce^{kt}$

$31.364 - 70 = Ce^{-0.121\,227\,160\,7(0)}$

$C = -38.636^\circ F$

Therefore, the equation when the thermometer is inside the room is

$T - 70 = -38.636e^{-0.121\,227\,160\,7t}$

At 9:20 am, 10 minutes after the thermometer was taken indoors

$T - 70 = -38.636e^{-0.121\,227\,160\,7(10)}$

$T = 58.505^\circ F$

For

T= 70°F$70 - 70 = -38.636e^{-0.121\,227\,160\,7t}$

Since logarithm of zero do not exist, we can approximate the solution. Use

T→ 70^{-}, sayT= 69.99999 then solve fort. It will actually do the job.I leave it to you then.

## If I am not mistaken sir, "to

If I am not mistaken sir, "to the nearest degree" means that the temperature reading will be $69.5^{\circ} F$ ?

## I thought it was rounding off

I thought it was rounding off, hehehe.