#### 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

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.