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13/09/2021 4:09 pm
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A cup of coffee cools from 90°C to 80°C in t minutes, when the room temperature is 20°C. The time taken by a similar cup of coffee to cool from 80°C to 60°C at a room temperature same at 20°C is :
(1) \(\frac{13}{10}t\)
(2) \(\frac{13}{5}t\)
(3) \(\frac{10}{13}t\)
(4) \(\frac{5}{13}t\)
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13/09/2021 4:16 pm
Answer: (2) \(\frac{13}{5}t\)
Explanation:
According to Newton's law of cooling
\(\frac{T_1 - T_2}{t}\) = K\(\Big[\frac{T_1 + T_2}{2} - T_0 \Big]\)
For 1st cup of coffee
⇒ \(\frac{90 - 80}{t}\) = K\(\Big[\frac{90 + 80}{2} - 20\Big]\) .....(i)
For 2nd cup of coffee,
⇒ \(\frac{80 - 60}{t'}\) = K\(\Big[\frac{80 + 60}{2} - 20\Big]\) .....(ii)
Divided (i) by (ii)
\(\frac{t'}{2t}\) = \(\frac{65}{50}\) ⇒ t' = \(\frac{13}{5}t\)
This post was modified 3 years ago by admin