Properties Of Solids And Liquids Question 458
Question: A body cools in a surrounding of constant temperature $ 39{}^\circ C $ . Its heat capacity is$ 2J/kg {}^\circ C $ . Initial temperature of the body is$ 49{}^\circ C $ . Assume Newton’s law of cooling is valid. The body cools to $ 38{}^\circ C $ in 19 minutes. In further 19 minutes it will cool from $ 38{}^\circ C $ to
Options:
A) $ 36{}^\circ C $
B) $ 36.4{}^\circ C $
C) $ 37{}^\circ C $
D) $ 37.5{}^\circ C $
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Answer:
Correct Answer: B
Solution:
[b] We have$ \theta -{\theta _{s}}=({\theta _{0}}-{\theta _{s}}){{e}^{-kt}} $ ….(l)
where $ {\theta _{0}} $ = initial temperature of body $ =40{}^\circ C $
$ \theta = $ temperature of body after time t $ {\theta _{s}}= $ temperature of surrounding Since body cools from $ 40{}^\circ C to 38{}^\circ C $ in 10 min, we have $ 38{}^\circ -30{}^\circ = ( 40{}^\circ -30{}^\circ ) {{e}^{-10k}} $ ….(2)
Let after 10 min, the body temp. be $ \theta $ .
$ \theta -30{}^\circ =( 38{}^\circ -30{}^\circ ){{e}^{-10k}} $ …..(3)
Dividing equ. (2) by equ. (3) gives, $ \frac{{{8}^{o}}}{\theta -{{30}^{o}}}=\frac{{{10}^{o}}}{{{8}^{o}}}\Rightarrow \theta -{{30}^{o}}={{6.4}^{o}}\Rightarrow \theta =36.{{4}^{o}} $
