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# A Body Cools at the Rate of 0.5°C / Minute When It is 25° C Above the Surroundings. Calculate the Rate of Cooling When It is 15°C Above the Same Surroundings. - Physics

#### Question

A body cools at the rate of 0.5°C / minute when it is 25° C above the surroundings. Calculate the rate of cooling when it is 15°C above the same surroundings.

#### Solution

Given: theta_1=25^@C, theta_2=15^@C

[(d theta)/dt]=0.5^@C/min

To Find: Rate of cooling at theta_2((d theta)/dt)_2

(d theta)/dt=K(theta-theta_0)

Using formulae for theta_2=15^@C

((d theta)/dt)_2/((d theta)/dt)_1=(theta_2-theta_0)/(theta_1-theta_0)

((d theta)/dt)_2/0.5=15/25

=0.3^@C/min

The rate of cooling when it is 15°C above same surroundings is 0.3°C/min.

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#### APPEARS IN

2016-2017 (March) (with solutions)
Question 2.8 | 2.00 marks

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Solution A Body Cools at the Rate of 0.5°C / Minute When It is 25° C Above the Surroundings. Calculate the Rate of Cooling When It is 15°C Above the Same Surroundings. Concept: Derivation of Boyle’s Law.
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