Temperature Control in Cooking

Question

Temperature Control in Cooking

In cooking, understanding the cooling/warming rates of different foods is crucial. Consider a dish taken out of an oven with an initial temperature of T0=180∘C. Using the cooling constant k=−0.03, calculate the time it takes for the dish to cool down to a safe eating temperature of Tm=60∘C.

Answer 1

Temperature Control in Cooking

In cooking, understanding the cooling/warming rates of different foods is crucial. Consider a dish taken out of an oven with an initial temperature of T0=180∘C. Using the cooling constant k=−0.03, calculate the time it takes for the dish to cool down to a safe eating temperature of Tm=60∘C.

To solve this we can use use "Newton's Law of Cooling Formula"

T(t) = T_s + (T₀ - T_s) e^-kt

T(t) = temperature of an object at a certain time (Kelvin, K)

t = time (s)

T_s = temperature of the surroundings (Kelvin, K)

T₀ = starting temperature of the object (Kelvin, K)

k = a cooling constant, specific to the object (1/s)

Note Kelvin is C° + 273

Given:

T₀=180∘C The formula used Kelvin, so 180+273 = 453 K°

k = 0.03

T_m=60∘C 60 + 273 = 333 K°

Find:

The time is takes the dish to go from T₀ to T_m

Assumptions: We will assume the room temperature is 20°C (293K°) as it is not given in the problem. That will be T_s

Now we can substitute the information above into the equation and solve for t

T(t) = T_s + (T₀ - T_s) e^-kt

333 = 293 +(453-293)e^(-0.03)t

40 = 160 e^(-0.03)t

0.25 = e^(-0.03)t

ln(0.25) = -0.03 t

t = 46.21 seconds = 0.77 Minutes

Hope this helps :)