A pizza pan is removed at 900 PM from an oven whose temperat

A pizza pan is removed at 9:00 PM from an oven whose temperature is fixed at 400°F into a room that is a constant 75°F.

After 5 minutes, the pizza pan is at 300°F.

(a) At what time is the temperature of the pan 135°F?

(b) Determine the time that needs to elapse before the pan is 220°.

The pizza will obey Newton\'s Law of cooling.

Therefore its temperature at time t will be given by: T(t) = T_s + (T_i - T_s)e^-kt

where T_s is the temperature of the surroundings and T_i is the initial temperature.

T(t) = 300 F = 148.88^o C

T_s = 75 F = 23.88^oC

T_i = 400 F = 204.44^oC

Therefore, substituting for t, T(t) T_s and T_i we get:

148.88 = 23.88 + (204.44 - 23.88)e^-k(5)

(204.44 - 23.88)e^-k(5) = 125

=> e^-k(5) = 0.6923

taking ln on both sides, we get:

- k(5) = ln(0.6923)

therefore k = 0.07354 min^-1

a] T(t) = 135 F = 57.22^oC

T_s = 75 F = 23.88^oC

T_i = 400 F = 204.44^oC

Therefore, substituting for t, T(t) T_s and T_i we get:

57.22 = 23.88 + (204.44 - 23.88)e^-kt

(204.44 - 23.88)e^-kt = 33.33

=> e^-kt = 0.1846

taking ln on both sides and substituting for k = 0.07354, we get:

- (0.07354) t = ln(0.1846)

=> t = 22.97 mins or 23 minutes.

b] T(t) = 220 F = 104.44^oC

substitute for other values, to get

104.44 = 23.88 + (204.44 - 23.88)e^-kt

(204.44 - 23.88)e^-kt = 80.55

=> e^-kt = 0.44615

taking ln on both sides and substituting for k = 0.07354, we get:

- (0.07354) t = ln(0.44615)

=> t = 10.97 mins or 11 minutes.

c] As time passes, one notices that the rate of cooling decreases exponentially which means that the pizza takes more time to cool.