Question

The time taken for a substance to cool form 100° C to 50° C is 5 minutes. If the room temperature is 30°C, what time required for same substance to cool from 50°C to 40°C?

Answer 1

Case (i):

Rate of cooling

\(\frac{dθ}{dt} \) = \(\frac {100-50}{2}\) = \(10°\)C/min,

Average temperature

θ = \(\frac {100+50}{2}\) = 75°C

Room temperate θ₀ = 30°C.

According to Newton’s law of cooling,

\(\frac{dθ }{dt}\) = -K(θ – θ₀).

10 = -K(75 – 30)

K= \(-\frac{10}{45}\)/min ....(1)

Case(ii):

Rate of cooling

Average temperature

Room temperate e θ₀ = 30°C.

According to Newton’s law of cooling,

\(\frac{dθ }{dt}\) = K (θ - θ₀)

\(\frac{10}{dt}\) = -(-10/45) (45-30) (from)

dt = 3 min.