A) 372, 310
B) 472, 410
C) 310, 372
D) 744, 682
Correct Answer: A
Solution :
: In first case, \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{output}{input}\] \[\therefore \] \[\frac{1}{6}=1-\frac{{{T}_{2}}}{{{T}_{1}}}\] \[\therefore \] \[\frac{{{T}_{2}}}{{{T}_{1}}}=1-\frac{1}{6}=\frac{5}{6}\]or\[{{T}_{1}}=\frac{6}{5}{{T}_{2}}=1.2{{T}_{2}}\]?(i) In second case, \[\eta =1-\frac{{{T}_{2}}-62}{{{T}_{1}}};\] \[\frac{1}{3}=1-\frac{{{T}_{2}}-62}{{{T}_{1}}}\] \[\frac{{{T}_{2}}-62}{{{T}_{1}}}=1-\frac{1}{3}=\frac{2}{3};\] \[\frac{({{T}_{2}}-62)}{1.2{{T}_{2}}}=\frac{2}{3}\] or \[3{{T}_{2}}-186=2.4{{T}_{2}}\] or \[0.6{{T}_{2}}=186\] or \[{{T}_{2}}=\frac{186}{0.6}=310\,K\] Hence \[{{T}_{1}}=1.2{{T}_{2}}=1.2\times 310=372\text{ }K\] Source\[=372\text{ }K\] Sink\[=310\text{ }K\].
You need to login to perform this action.
You will be redirected in
3 sec
