A cake is removed from a 375°F oven and placed on a cooling rack in a 63°F room. After 30 minutes the cake is 175°F. When will it be 150°F? (Round your answer to two decimal places.)

The cake will be at temperature 150°F at after 37.34 minutes

Explanation:

Let T be the temperature of the cake at any time

T∞ be the temperature around the cooling rack = 63°F

T₀ be the initial temperature of the cake = 375°F

And m, c, h are all constants from the cooling law relation

From Newton's law of cooling

Rate of Heat loss by the cake = Rate of Heat gain by the environment

- mc (d/dt)(T - T∞) = h (T - T∞)

(d/dt) (T - T∞) = dT/dt (Because T∞ is a constant)

dT/dt = (-h/mc) (T - T∞)

Let (h/mc) be k

dT/(T - T∞) = -kdt

Integrating the left hand side from T₀ to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -kt

(T - T∞)/(T₀ - T∞) = e⁻ᵏᵗ

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

Inserting the known variables

(T - 63) = (375 - 63)e⁻ᵏᵗ

(T - 63) = 312 e⁻ᵏᵗ

At 30 minutes, T = 175°F

175 - 63 = 312 e⁻ᵏᵗ

112/312 = e⁻ᵏᵗ

- kt = In (112/312) = In (0.3590)

- 30k = - 1.025

k = 1.025/30 = 0.0342 /min

When the temp is 150°F,

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

(150 - 63) = 312 e⁻ᵏᵗ

e⁻ᵏᵗ = (87/312) = 0.2788

- kt = In 0.2788 = - 1.277

t = 1.277/k = 1.277/0.0342 = 37.34 min.

the duration of a photographic flash is related to an rc time constant, which is 0.100 μs for a certain camera.