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11.Thermodynamics
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ચાનો કપ $80° C$ થી $60° C$ એ એક મિનિટમાં ઠંડો પડે છે. ન્યૂનત્તમ તાપમાન $ 30° C$ છે. તો$60° C $થી $50° C$ ઠંડો થવા ...... $\sec$ સમય લાગશે?
A
$50$
B
$90$
C
$60$
D
$48$
Solution
$\frac{{{\theta _{\text{1}}} – {\theta _2}}}{t} = K\left[ {\frac{{{\theta _1} + {\theta _2}}}{2} – {\theta _0}} \right]$
$\frac{{80 – 60}}{{60}} = K\left[ {\frac{{80 + 60}}{2} – 30} \right]\,\,\, \Rightarrow \,\,\,\frac{{20}}{{60}} = K[70 – 30] = 40K$
$K = \frac{1}{{120}}\,\,\,\,\frac{{60 – 50}}{t} = \frac{1}{{120}}\left[ {\frac{{110}}{2} – 30} \right]$
$\frac{{10}}{t} = \frac{1}{{120}}[25] \Rightarrow \,t = \frac{{1200}}{{25}} = 48\,\,\sec $
Standard 11
Physics
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