Question

# The temperature of a body falls from 40∘C to 36∘C in 5 minutes when placed in a surrounding of constant temperature 16∘C. Find the time taken for the temperature of the body to become 32∘C.8.1 min 6.1 min 5.1 min 5.3 min

Solution

## The correct option is B 6.1 min As the temperature differences are small, we can use Newton's law of cooling. dθdt=−kt(θ−θ0) dθθ−θ0=−kdt or,                                               ...(i) Where k is a constant, θ is the temperature of the body at time t and θ0=16∘C is the temperature of the surrounding. We have, ∫36∘C40∘Cdθθ−θ0=−kt(5 min) ln36∘C−16∘C40∘C−16∘C=−kt(5 min) or,                                      k=−ln(56)5 min or, . If t be the time required for the temperature to fall from 36∘C to 32∘C then by (i), ∫32∘C36∘Cdθθ−θ0=−kt or,        ln32∘C−16∘C36∘C−16∘C=−ln(56)5 min or,          t=ln(45)(ln(56)×5min = 6.1 min. Alternative method The mean temperature of the body as it cools from 40∘C to 36∘C is  40∘C+36∘C2=38∘C The rate of decrease of temperature is   40∘C+36∘C5min=0.80∘C/min . Newton's law of cooling is dθdt=−k(θ−θ0) or, -0.8∘C/min = - k(38∘C - 16∘C)=-k(22∘C) or, k=0.822min−1. Let the time taken for the temperature to become 32∘C be t. During this period,. The mean temperature is Now,

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