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  • How to solve this problem- - Properties of Solids and Liquids - JEE Main

A body takes 10 minutes to cool from 60^{\circ} C to 50^{\circ} C. The temperature of surroundings is constant at 25^{\circ} C. Then, the temperature of the body after next 10 minutes will be approximately :

  • Option 1)

    47^{\circ} C

  • Option 2)

    41^{\circ} C

  • Option 3)

    45^{\circ} C

  • Option 4)

    43^{\circ} C^{}

 

Answers (2)

best_answer

As we learnt

When body Cools by Radiation from theta1 degree c to theta two degree c in time t -

\left[\frac{\theta_{1}-\theta_{2}}{t} \right ]=k\left[\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0} \right ]

- wherein

\theta_{av}=\frac{\theta_{1}+\theta_{2}}{2}

 

 \frac{T_{2}-T_{1}}{t}=K.\left [\frac{T_{1}+T_{2}}{2} -T_{0} \right ]

\therefore \frac{60-50}{10}=K.\left [ 55-25 \right ]\rightarrow (1)

\therefore \frac{50-T}{10}=K.\left [ \frac{50+T}{2} -25\right ]

Divide both eqns

\therefore \frac{10}{50-T}=\frac{30}{\left ( \frac{T}{2} \right )}= \frac{60}{T}

\therefore T=300-6T\: \: or\: \: 7T=300

\therefore T=\frac{300}{7}=43^{\circ}C


Option 1)

47^{\circ} C

Option 2)

41^{\circ} C

Option 3)

45^{\circ} C

Option 4)

43^{\circ} C^{}

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Plabita

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