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Properties of Matter question

2014 · Q153
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Properties of Matter question

2014 · Q153

NEETPhysicsProperties of MatterMCQ+4 / −1
Certain quantity of water cools from 70oC to 60oC in the first 5 minutes and to 54oC in the next 5 minutes. The temperature of the surroundings is
  1. A
    45oC
  2. B
    20oC
  3. C
    42oC
  4. D
    10oC
View written solutionFree

Correct answer: A

Let the temperature of surroundings be θ0{\theta _0}θ0​
By Newton's law of cooling

θ1−θ2t=k[θ1−θ22−θ0]{{{\theta _1} - {\theta _2}} \over t} = k\left[ {{{{\theta _1} - {\theta _2}} \over 2} - {\theta _0}} \right]tθ1​−θ2​​=k[2θ1​−θ2​​−θ0​]

⇒70−605=k[70+602−θ0] \Rightarrow {{70 - 60} \over 5} = k\left[ {{{70 + 60} \over 2} - {\theta _0}} \right]⇒570−60​=k[270+60​−θ0​]

⇒2=k[65−θ0] \Rightarrow 2 = k\left[ {65 - {\theta _0}} \right]⇒2=k[65−θ0​]

Similarly, 60−545=k[60+542−θ0]{{60 - 54} \over 5} = k\left[ {{{60 + 54} \over 2} - {\theta _0}} \right]560−54​=k[260+54​−θ0​]

⇒65=k[57−θ0] \Rightarrow {6 \over 5} = k\left[ {57 - {\theta _0}} \right]⇒56​=k[57−θ0​]

By dividing (i) by (ii) we have

106=65−θ057−θ0⇒θ0=45o{{10} \over 6} = {{65 - {\theta _0}} \over {57 - {\theta _0}}} \Rightarrow {\theta _0} = {45^o}610​=57−θ0​65−θ0​​⇒θ0​=45o

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