Hot water cools from {60^o}C to {50^o}C in the first 10 minutes and to {42^o}C in the next 10 minutes. The temperature of the surrounding

Q: Hot water cools from ${60^o}C$ to ${50^o}C$ in the first $10$ minutes and to ${42^o}C$ in the next $10$ minutes. The temperature of the surrounding is ......... $^oC$

b (b) According to Newton's law of cooling $\frac{{{\theta _1} - {\theta _2}}}{t} = K\left[ {\frac{{{\theta _1} + {\theta _2}}}{2} - {\theta _0}} \right]$ In the first case, $\frac{{(60 - 50)}}{{10}} = K\,\left[ {\frac{{60 + 50}}{2} - {\theta _0}} \right]$ $1 = K\,(55 - \theta )$….$(i)$ In the second case, $\frac{{(50 - 42)}}{{10}} = K\,\left[ {\frac{{50 + 42}}{2} - {\theta _0}} \right]$ $0.8 = k\,(46 - {\theta _0})$….$(ii)$ Dividing $(i)$ by $ (ii)$ , we get $\frac{1}{{0.8}} = \frac{{55 - \theta }}{{46 - \theta }}$ or $46 - {\theta _0} = 44 - 0.8\theta $ ==> ${\theta _0} = {10^o}C$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

Hot water cools from ${60^o}C$ to ${50^o}C$ in the first $10$ minutes and to ${42^o}C$ in the next $10$ minutes. The temperature of the surrounding is ......... $^oC$

A$5$
B$10$
C$15$
D$20$

Diffcult

STD 11 Science
NEET
STD 11 - 10.2. transmission of heat
Physics

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