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Heat and Thermodynamics question

2021 · Q153
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Heat and Thermodynamics question

2021 · Q153

NEETPhysicsHeat and ThermodynamicsMCQ+4 / −1
A cup of coffee cools from 90∘^\circ∘C to 80∘^\circ∘C in t minutes, when the room temperature is 20∘^\circ∘C. The time taken by a similar cup of coffee to cool from 80∘^\circ∘C to 60∘^\circ∘C at a room temperature same at 20∘^\circ∘C is :
  1. A
    513{5 \over {13}}135​t
  2. B
    1310{13 \over {10}}1013​t
  3. C
    135{13 \over {5}}513​t
  4. D
    1013{10 \over {13}}1310​t
View written solutionFree

Correct answer: C

dTdt=K(Tav−T0){{dT} \over {dt}} = K({T_{av}} - {T_0})dtdT​=K(Tav​−T0​)

10t=K(85−20){{10} \over t} = K(85 - 20)t10​=K(85−20)

20t=K(70−20){{20} \over t} = K(70 - 20)t20​=K(70−20)

t′2t=6550{{t'} \over {2t}} = {{65} \over {50}}2tt′​=5065​

t′t=13050{{t'} \over t} = {{130} \over {50}}tt′​=50130​

t′=135tt' = {{13} \over 5}tt′=513​t

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