A) \[30.5{}^\circ C\]
B) \[37{}^\circ C\]
C) \[27{}^\circ C\]
D) \[32{}^\circ C\]
Correct Answer: D
Solution :
[d] Both oxygen and nitrogen will have same specific heat as they are diatomic gases. Let \[\theta \] be the temp. of mixture. Loss of heat = gain of heat \[\frac{16}{16}\times {{C}_{p}}\times (37-\theta )=\frac{14}{44}\times {{C}_{p}}(\theta -27)\] \[2\theta =37+27=64\]\[\Rightarrow \]\[\theta =\frac{64}{2}=32{}^\circ C.\]
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