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A body cools in 7 minutes from  60^{\circ} \mathrm{C}  to  40^{\circ} \mathrm{C} .  The temperature of the surrounding is 10^{\circ} \mathrm{C} . The temperature of the body after the next 7 minutes

Option: 1

30^{\circ} \mathrm{C}


Option: 2

34^{\circ} \mathrm{C}


Option: 3

32^{\circ} \mathrm{C}


Option: 4

28^{\circ} \mathrm{C}


Answers (1)

best_answer

Method-1
Using exact law of cooling

\mathrm{T}-\mathrm{T}_{\mathrm{s}}=\left(\mathrm{T}_0-\mathrm{T}_{\mathrm{s}}\right) \mathrm{e}^{-\mathrm{Kt}}

Case-I:  (40-10)=(60-10) \mathrm{e}^{-7 \mathrm{~K}}

30=50e^{-7k} \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ..... (1)

Case-II: (T-10)=(40-10)e^{-7k}    or  T-10=30e^{-7K}

Dividing (2) by (1)

\begin{aligned} & \frac{\mathrm{T}-10}{30}=\frac{30}{50} \\ & \Rightarrow \mathrm{T}-10=\frac{30 \times 30}{50}=18 \end{aligned}

Or  T=28^{\circ} \mathrm{C}

Methode-2
Using newton’s average law of cooling

\frac{T_i-T_f}{t}=k\left(\frac{T_i+T_f}{2}-T_s\right)

  Case-I:-     \frac{60-40}{7}=\mathrm{R}\left[\frac{60+40}{2}-10\right] \Rightarrow \frac{20}{7}=\mathrm{k}[40]                         .......(i)

 Case-II:- Case-II:- \frac{40-\mathrm{T}}{7}=\mathrm{R}\left[\frac{20+\mathrm{T}}{2}\right]                                          ........(ii)

Dividing (2) by (1)

\begin{aligned} & \frac{40-T}{20}=\frac{20+T}{80} \\ & 160-4 \mathrm{~T}=20+\mathrm{T} \\ & 5 \mathrm{~T}=140 \\ & \mathrm{~T}=28^{\circ} \mathrm{C} \end{aligned} 

Posted by

avinash.dongre

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