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  •  Hot water cools from 600C to 500C in the first 10 minutes and to 420C in the next  10 minutes. The temperature of the  surroundings is :  

 Hot water cools from 600C to 500C in the first 10 minutes and to 420C in the next  10 minutes. The temperature of the  surroundings is :        

 

Option: 1

250C


Option: 2

100C


Option: 3

150C


Option: 4

200C


Answers (1)

best_answer

From Newton's Law of Cooling,  we know that

 When body Cools by Radiation from   \ \theta_1^0 C to theta \ \theta_2^0 C in time t                                                                                                                                     

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

So

where $\theta_{0}$ is the temperature of the surrounding.

Now, the hot water cools from $60^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in 10 minutes
\frac{60-50}{10}=-\mathrm{K}\left[\frac{60+50}{2}-\theta_{0}\right]$.....(i)
Again, it cools from $50^{\circ} \mathrm{C}$  to $42^{\circ} \mathrm{C}$ in next 10 minutes.
\frac{50-42}{10}=-\mathrm{K}\left[\frac{50+42}{2}-\theta_{0}\right]$......(ii)

Dividing equations (i) by (ii) we get
\frac{10}{8}=\frac{55-\theta_{0}}{46-\theta_{0}}$\\\\ $460-10 \theta_{0}=440-8 \theta_{0}$\\ $2 \theta_{0}=20$\\ $\theta_{0}=10^{\circ} \mathrm{C}$

Posted by

Gautam harsolia

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