A) \[\frac{({{T}_{1}}+{{T}_{2}}+{{T}_{3}})}{3}\]
B) \[\frac{{{n}_{1}}{{T}_{1}}+{{n}_{2}}{{T}_{2}}+{{n}_{3}}{{T}_{3}}}{{{n}_{1}}+{{n}_{2}}+{{n}_{3}}}\]
C) \[\frac{{{n}_{1}}{{T}_{1}}+{{n}_{2}}T_{2}^{2}+{{n}_{3}}T_{3}^{2}}{{{n}_{1}}{{T}_{1}}+{{n}_{2}}{{T}_{2}}+{{n}_{3}}{{T}_{3}}}\]
D) \[\frac{n_{1}^{2}T_{1}^{2}+n_{3}^{2}T_{2}^{2}+n_{3}^{2}T_{3}^{2}}{{{n}_{1}}{{T}_{1}}+{{n}_{2}}{{T}_{2}}+{{n}_{3}}{{T}_{3}}}\]
Correct Answer: B
Solution :
| [b] Let \[{{T}_{3}}>{{T}_{2}}>{{T}_{1}}\] and final temperature is T such that \[{{T}_{3}}>T>{{T}_{2}}>{{T}_{1}}\]. Now heat gained by first tow gases is equal to heat lost by third gas |
| \[{{n}_{1}}C(T-{{T}_{1}})+{{n}_{2}}C(T-{{T}_{2}})={{n}_{3}}C({{T}_{3}}-T)\] |
| \[\Rightarrow \] \[T=\frac{{{n}_{1}}{{T}_{1}}+{{n}_{2}}{{T}_{2}}+{{n}_{3}}{{T}_{3}}}{{{n}_{1}}+{{n}_{2}}+{{n}_{3}}}\] |
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