Question

A pizza pan is removed at 6:00 PM from an oven whose temperature is fixed at 425°F into a room that is a constant 71°F. After 5 minutes, the pizza pan is at 300°F. (a) At what time is the temperature of the pan 130°F? (b) Determine the time that needs to elapse before the pan is 210 (c) What do you notice about the temperature as time passes?

Answer 1

(a) To find the time when the temperature of the pan is 130°F, we can use the formula:

T(t) = T0 + (T1 - T0)e^(-kt)

where T(t) is the temperature of the pan at time t, T0 is the initial temperature of the pan, T1 is the ambient temperature, k is a constant, and e is the base of the natural logarithm.

We know that T0 = 425°F, T1 = 71°F, and T(t) = 130°F. We can solve for k by using the fact that after 5 minutes, the temperature of the pan is 300°F:

300 = 71 + (425 - 71)e^(-5k)

Solving for k, we get:

k = ln(354/354e^(-5k)) / (-5)

k = 0.2301

Substituting T0 = 425°F, T1 = 71°F, and k = 0.2301 into the formula, we get:

130 = 425 + (71 - 425)e^(-0.2301t)

Solving for t, we get:

t = 29.8 minutes

So, the temperature of the pan is 130°F at approximately 6:30 PM.

(b) To find the time that needs to elapse before the pan is 210°F, we can use the same formula:

210 = 425 + (71 - 425)e^(-0.2301t)

Solving for t, we get:

t = 47.8 minutes

So, the time that needs to elapse before the pan is 210°F is approximately 7:00 PM.

(c) As time passes, the temperature of the pan approaches the ambient temperature of 71°F. The rate at which the temperature of the pan changes is proportional to the difference between the temperature of the pan and the ambient temperature. This is why the temperature of the pan decreases rapidly at first and then more slowly as it approaches the ambient temperature.