HSC Science (General) 12th Board ExamMaharashtra State Board
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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? - HSC Science (General) 12th Board Exam - Mathematics and Statistics

ConceptDifferential Equations Applications of Differential Equation

Question

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 θ be the temperature of the body at any time t.
Temperature of air is given to be 10o C.
According to Newton's law of cooling, we have

(d theta)/dt prop theta-10^@

(d theta)/dt =-k( theta-10^@),k>0

(d theta)/(theta-10^@)=-kdt

Integrating both sides, we get

int(d theta)/(theta-10^@)=-kintdt

ln(theta-10^@)=-kt+C

theta=10^@+e^(-k(t)+c).........(1)

When t = 0, θ = 110o

Substituting in the equation, we get

110^@=10^@+e^(-k(0)+c)

e^c=100^@

Substituting the above in (1), we get

theta=10^@+100^@e^(-kt).........(2)

As per the data in the question,

60^@=10^@+100^@e^(-k(1))

50^@=100^@e^(-k(1)).........(3)

e^(-k/2)=1/2

k=ln2

35^@=10^@+100^@e^(-kt)

25^@=100^@e^(-kt)...........(4)

Dividing (4) by (3), we get

2=e^(-k(1-t))

ln2=-k(1-t)

t-1=ln2/k=ln2/ln2=1

Hence, additional time required for cooling from 60o to 35o is 1 hour.

Is there an error in this question or solution?

APPEARS IN

2015-2016 (March) (with solutions)
Question 5.2.3 | 4.00 marks

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Solution 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? Concept: Differential Equations - Applications of Differential Equation.
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