Answer:
a) The reading of the thermometer at t = 1 min is 36.67°F
b) Time it will take for the thermometer to reach 15° F = 3.06 minutes
Explanation:
Let T be the temperature of the thermometer at any time
T∞ be the temperature of the air = 10°F
T₀ be the initial temperature of the thermometer = 70°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 - 10) = (70 - 10)e⁻ᵏᵗ
(T - 10) = 60 e⁻ᵏᵗ
At t = 0.5 minute, T = 50°F
50 - 10 = 60 e⁻ᵏᵗ
40/60 = e⁻ᵏᵗ
- kt = In (40/60) = In (0.6667)
- 0.5k = - 0.4054
k = 0.811 /min
At time t = 1 min
kt = 0.811 × 1 = 0.811
(T - 10) = 60 e⁻ᵏᵗ
e^(-0.811) = 0.4445
T - 10 = 60 × 0.4445 = 26.67
T = 36.67°F
b) When the temp is 15°F,
(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ
(15 - 10) = 60 e⁻ᵏᵗ
e⁻ᵏᵗ = (5/60) = 0.08333
- kt = In 0.0833 = - 2.485
t = 2.485/k = 2.485/0.811 = 3.06 min.
