Question

An object is taken from an oven at 350o F and left to cool in a room at 70o F. If the temperature fell to 250o F in one hour, what would its temperature be 4 hours after it was removed from the oven

Answer 1

Answer:

117.83° F

Explanation:

Using Newton's Law of Cooling which can be expressed as:

$\dfrac{dT}{dt}= k(T-T_1)$

The differential equation can be computed as:

$\dfrac{dT}{dt}= k(T-70)$

$\dfrac{dT}{(T-70)}= kdt$

$\int \dfrac{dT}{(T-70)}= \int kdt$

$In|T-70| = kt +C$

$T- 70 = e^{kt+C} \\ \\ T = 70+e^{kt+C} \\ \\ T = 70 + C_1e^{kt} --- (1)$

where;

$C_1 = e^C$

At the initial condition, T(0)= 350

$350 = 70 C_1^{k*0}$

$350 -70 = C_1$

$280 = C_1$

replacing $C_1$= 280 into (1)

Hence, the differential equation becomes:

$T(t) = 70 + 280 e^{kt}$

when;

time (t) = 1 hour

T(1) = 250

Since;

$250 = 70 + 280 e^{k*1}$

$180 = 280e^k \\ \\ \dfrac{180}{280}= e^k$

$k = In (\dfrac{180}{280})$

k = -0.4418

Therefore;

T(t) = 70 + 280e^{(-0.4418)}t

After 4 hours, the temperature is:

T(t) = 70 + 280e^{(-0.4418)}4

T(4) = 117.83° F