A) \[45{}^\circ C\]
B) \[20{}^\circ C\]
C) \[42{}^\circ C\]
D) \[10{}^\circ C\]
Correct Answer: A
Solution :
| Let the temperature of the surrounding is \[t{}^\circ C\]. |
| For first case, |
| \[\frac{(70-60)}{5\min }=k({{65}^{{}^\circ }}C-{{t}^{{}^\circ }}C)\] |
| (\[65{}^\circ \] is average of \[70{}^\circ C\] and \[60{}^\circ C\]) |
| \[\frac{10}{5\min }=K({{65}^{{}^\circ }}C-{{t}^{{}^\circ }}C)\] (i) |
| For second case, |
| \[\frac{(60-54)}{5\min }=k(57-t)\] (ii) |
| (\[57{}^\circ C\] is average of \[60{}^\circ C\] and \[54{}^\circ C\]) |
| From Eqs. (i)/(ii), \[\frac{10}{6}=\frac{(65-t)}{(57-t)}\] |
| Solving, we get \[t=45{}^\circ C\] |
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