A thermometer initially reading 212F is placed in a room where the temperature is 70F. After 2 minutes the thermometer reads 125F. (a) What does the thermometer read after 4 minutes

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