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

117.83° F

Explanation:

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

The differential equation can be computed as:

where;

At the initial condition, T(0)= 350

replacing = 280 into (1)

Hence, the differential equation becomes:

when;

time (t) = 1 hour

T(1) = 250

Since;

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