Transmission Of Heat Question 165
Question: Hot water cools from $ 60^{o}C $ to $ 50^{o}C $ in the first 10 minutes and to $ A $ in the next 10 minutes. The temperature of the surrounding is
[MP PET 1993]
Options:
A) $ 5^{o}C $
B) $ 10^{o}C $
C) $ 15^{o}C $
D) $ 20^{o}C $
Show Answer
Answer:
Correct Answer: B
Solution:
According to Newton’s law of cooling $ \frac{{\theta _{1}}-{\theta _{2}}}{t}=K[ \frac{{\theta _{1}}+{\theta _{2}}}{2}-{\theta _{0}} ] $ In the first case, $ \frac{(60-50)}{10}=K,[ \frac{60+50}{2}-{\theta _{0}} ] $
$ 1=K,(55-\theta ) $ ?.(i) In the second case, $ \frac{(50-42)}{10}=K,[ \frac{50+42}{2}-{\theta _{0}} ] $
$ 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 $
Therefore $ {\theta _{0}}=10^{o}C $
BETA
