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BITSAT 2011 Paper

Section: Physics

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Question : 15 of 150

Marks: +1, -0

A copper sphere cools from 62°C to 50°C in 10 minutes and to 42°C in the next 10 minutes. Calculate the temperature of the surroundings.

18.01ºC
26ºC
10.6ºC
20ºC

Solution:

By Newton's law of cooling,

\frac{\theta_1-\theta_2}{t}

= - k

\left[\frac{\theta_1+\theta_2}{2} - \theta_0\right]

... (1)

A sphere cools from 62°C to 50°C in 10 min.

\frac{62-50}{10}

= - k

\left[\frac{62+50}{2} - \theta_0\right]

... (2)

Now, sphere cools from 50°C to 42°C in next 10 min.

\frac{50-42}{10}

= - k

\left[\frac{50+42}{2} - \theta_0\right]

... (3)

Dividing eqn. (2) by (3) we get,

\frac{56-\theta_0}{46-\theta_0}

or

0.4\theta_0

= 10.4

Hence

\theta_0

= 26°C

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