# A Body is Heated at 110°C and Placed in Air at 10°C. After 1 Hour Its Temperature is 60°C. How Much Additional Time is Required for It to Cool to 35°C? - Mathematics and Statistics

Sum

A body is heated at 110°C and placed in air at 10°C. After 1 hour its temperature is 60°C. How much additional time is required for it to cool to 35°C?

#### Solution

Let theta^circ "C" he the temperature of the body at any time t.
Temperature of air is 10o C i.e theta_circ = 10
According to Newton's law of cooling, we have

(d theta)/dt prop theta-theta_@

therefore (d theta)/dt =-k( theta-theta_@),k>0

therefore ("d"theta)/"dt" = - "K" (theta - 10)

therefore 1/(theta-10) "d"theta =-kdt

Integrating both sides, we get

int 1/(theta-10) "d"theta=- "k" int "dt"

therefore "log" (theta - 10) = -"Kt" + "C"

When t = 0 , θ = 110° C

∴ log (110 - 10) = - K × 0 + C        ∴ C = log 100

∴ log (θ - 10) = - Kt + log 100

"log" ((theta - 10)/100) = - "Kt"

Also θ = 60° C ; when t = 1

therefore "log"  ((60-10)/100) = - "K" xx 1

therefore "K" = - "log" (1/2)

therefore "log" ((theta - 10)/100) = "t"  "log" (1/2)

when θ = 35° C , then

"log" ((35-10)/100) = "t"  "log" (1/2)

"log" (25/100) = "t log" (1/2)

"log" (1/4) = "t log" (1/2)

- log 4 = - t log 2

t = ("log" 4)/("log" 2) = 2

The additional time required for body to cool to 35° C = (2 - 1) = 1 hour

Concept: Differential Equations - Applications of Differential Equation
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