JEE Mains · Physics · STD 11 - 10.2 transmission of heat
Hot water cools from \(60\,^oC\) to \(50\,^oC\) in the first \(10\,minutes\) and to \(42\,^oC\) in the next \(10\,minutes.\) The temperature of the surroundings is ...... \(^oC\)
- A \(25\)
- B \(10\)
- C \(15\)
- D \(20\)
Answer & Solution
Correct Answer
(B) \(10\)
Step-by-step Solution
Detailed explanation
By \(Newton's\,law\) of cooling \(\frac{{{\theta _1} - {\theta _2}}}{t} = - K\left[ {\frac{{{\theta _1} + {\theta _2}}}{2} - {\theta _0}} \right]\) where \({\theta _0}\) is the temperature of surrounding. Now, hot water cools from \({60^ \circ }C\,to\,{50^ \circ }C\) in…
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