A) \[25{}^\circ C\]
B) \[10{}^\circ C\]
C) \[15{}^\circ C\]
D) \[20{}^\circ C\]
Correct Answer: B
Solution :
[b] 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 10 minutes, \[\frac{60-50}{10}=-K\left[ \frac{60+50}{2}-{{\theta }_{0}} \right]\] ?(i) Again, it cools from \[50{}^\circ C\] to \[42{}^\circ C\] in next 10 minutes. \[\frac{50-42}{10}=-K\left[ \frac{50+42}{2}-{{\theta }_{0}} \right]\] ?(i) Dividing equations (i) by (ii) we get \[\frac{1}{0.8}=\frac{55-{{\theta }_{0}}}{46-{{\theta }_{0}}}\] \[\frac{10}{8}=\frac{55-{{\theta }_{0}}}{46-{{\theta }_{0}}}\] \[460-10{{\theta }_{0}}=440-8{{\theta }_{0}}\] \[2{{\theta }_{0}}=20\] \[{{\theta }_{0}}=10{}^\circ C\]
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