Answer to Question #267785 in Differential Equations for sebby

Question

A body of unknown temperature is placed in a refrigerator at a constant temperature of 0∘F, If after 20 minutes the temperature of the body is 40∘F, and after 40 minutes the temperature of the body is 20∘F, find the initial temperature of the body.

Answer 1

$\frac{dT}{dt} =- k(T-T_{s})$

Where

T = temperature of the body at time t

T_s = temperature of surrounding

k = Positive constant depends on the nature and area of the body surface under consideration

Here the surrounding temperature is 0⁰F that means T_s= 0

So $\frac{dT}{dt}=-kT$

So $\frac{dT}{T}=-kdt$

Integrating we get

$\int$ $\frac{dT}{T} =- \int kdt$

ln(T) = - kt + C where C is integration constant.

When t = 20 minutes, T = 40⁰F

So ln(40) = C - 20k •••••••equation(1)

When t = 40 minutes , T = 20⁰F

So ln(20) = C - 40k •••••••equation(2)

Multiplying equation (1) by 2 we get

2 ln(40) = 2C - 40k ••••••••equation(3)

and then subtracting equation (2) from equation (3) we get

2 ln(40) - ln(20) = C

=> ln(40)² - ln(20) = C

=> C = ln($\frac{(40)^{2}}{20}) = ln(80)$

So the Newtown's law of cooling becomes ln(T) = - kt + ln(80)

When t = 0 , T = T₀ , the initial temperature.

Therefore ln(T₀) = ln(80)

=> T₀ = 80

So initial temperature was 80⁰ F