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CGPET 2016 Solved Paper

Section: Physics

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Question : 6 of 150

Marks: +1, -0

A body initially at $80^{\circ}\mathrm{C}$ cools to $64^{\circ}\mathrm{C}$ in 5 minand to $52^{\circ}\mathrm{C}$ in 10 min. The temperature ofthis body after 15 min will be(assuming heat loss by radiation only)

40°C
43°C
41°C
39°C

Solution:

Let the temperature of surroundings is

\theta_0

Making use of Newton’s law of cooling,

\frac{\log(64- \theta_0)}{\log(80- \theta_0)} = -\frac{k}{m\times s} \times 5

...(i)

\log\left(\frac{64-\theta_0}{52-\theta_0}\right) = \frac{k}{m\times s} \times 5

In next 10 min, the temperature becomes

52

°.

\frac{\log(52- \theta_0)}{\log(64- \theta_0)} = -\frac{k}{m\times s} \times 10

...(ii)

On dividing Eq. (ii) by Eq. (i), we get

\log\left(\frac{52-\theta_0}{64-\theta_0}\right) = 2\log\left(\frac{64-\theta_0}{80-\theta_0}\right)

(52-\theta_0) \times (80-\theta_0) = (64-\theta_0)^2

or,

52 \times 80 - 132\theta_0 + \theta_0^2 = (64)^2 - 128\theta_0 + \theta_0^2

4\theta_0 = 4160 - 4096 = 64

\theta_0 = \frac{64}{4} = 16^{\circ}\mathrm{C}

\log\left(\frac{\theta-\theta_0}{80-\theta_0}\right) = \log\left(\left(\frac{64-16}{80-16}\right)^2\right)

where,

\theta

is the temperature of the body after 15 min.

\frac{\theta-16}{80-16} = \frac{27}{64}

\theta = 43^{\circ}\mathrm{C}

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