Respuesta :
Answer:
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.
