Answer:
91.3°F
Explanation:
Let T be the temperature of the thermometer at any time
T∞ be the temperature of the room = 70°F
T₀ be the initial temperature of the thermometer = 212°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 - 70) = (212 - 70)e⁻ᵏᵗ
(T - 70) = 142 e⁻ᵏᵗ
At t = 2 minute, T = 125°F
125 - 70 = 142 e⁻ᵏᵗ
55/142 = e⁻ᵏᵗ
- kt = In (55/142) = In (0.3873)
- k(2) = - 0.9485
k = 0.4742 /min
At time t = 4 mins
kt = 0.4742 × 4 = 1.897
(T - 70) = 142 e⁻ᵏᵗ
e^(-1.897) = 0.15
T - 70 = 142 × 0.15 = 21.3
T = 91.3°F
