A) \[10{}^\circ C\]
B) \[5{}^\circ C\]
C) \[15{}^\circ C\]
D) \[20{}^\circ C\]
E) \[22{}^\circ C\]
Correct Answer: A
Solution :
According to Newtons law of cooling \[\frac{{{\theta }_{1}}-{{\theta }_{2}}}{t}=K\left[ \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}-{{\theta }_{0}} \right]\] In the first case \[\Rightarrow \] \[\frac{60-50}{10}=K\left[ \frac{60+50}{2}-{{\theta }_{0}} \right]\] \[\Rightarrow \] \[1=K(55-\theta )\] ...(i) In the second case \[\Rightarrow \] \[\frac{50-42}{10}=K\left[ \frac{50+42}{2}-\theta \right]\] \[\Rightarrow \] \[0.8=K[46-\theta ]\] ...(ii) Dividing Eq. (i) by Eq. (ii), we get \[\frac{1}{0.8}=\frac{55-\theta }{46-\theta }\] or \[46-\theta =44-0.8\theta \] \[\theta =10{}^\circ C\]
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