Answer:
The temperature of the object at any time t, T(t) is given as
T = T∞ + (T₀ - T∞)e⁻⁰•⁵⁷²⁵ᵗ
Explanation:
Let T be the temperature of the object at any time
T∞ be the temperature outside = 30°
T₀ be the initial temperature of the object in the room = 69°
And m, c, h are all constants from the cooling law relation
From Newton's law of cooling
Rate of Heat loss by the object = Rate of Heat gain by the outside air
- 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 - 30) = (69 - 30)e⁻ᵏᵗ
(T - 30) = 39 e⁻ᵏᵗ
At 1 minute, T = 52°
52 - 30 = 39 e⁻ᵏᵗ
22/39 = e⁻ᵏᵗ
- kt = In (22/39) = In (0.564)
- k(1) = - 0.5725
k = 0.5725 /min
(T - T∞) = (T₀ - T∞)e⁻⁰•⁵⁷²⁵ᵗ
T = T∞ + (T₀ - T∞)e⁻⁰•⁵⁷²⁵ᵗ
