A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. After one-half minute the thermometer reads 50° F. What is the reading of the thermometer at t 1 min? How long will it take for the thermometer to reach 15° F?

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.

answer:

yes that is right have a good day