| \[{{Q}_{1}}=6000\text{ }J,\] | \[{{Q}_{2}}=-5500\text{ }J\] |
| \[{{Q}_{3}}=-3000\text{ }J,\] | \[{{Q}_{4}}=3500\text{ }J\] |
| \[{{W}_{1}}=2500\text{ }J,\] | \[{{W}_{2}}=-1000\text{ }J\] |
| \[{{W}_{3}}=-1200\text{ }J,\] | \[{{W}_{4}}=\times J\] |
A) 500, 7.5
B) 700, 10.5
C) 1000, 21
D) 1500, 15
Correct Answer: C
Solution :
According to the given problem \[\Delta Q={{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}+{{Q}_{4}}\] \[=\text{ }6000-5500-3000+3500\] \[\Delta Q=9500-8500=1000\text{ }J\] \[\Delta W={{W}_{1}}+{{W}_{2}}+{{W}_{3}}+{{W}_{4}}\] \[=2500\text{ }-1000\text{ }-\text{ }1200\text{ }+\text{ }x\] \[=\text{ }300\text{ }+\text{ }x\] And as for cyclic process, \[{{U}_{F}}={{U}_{1}},\Delta U={{U}_{F}}-{{U}_{I}}=0\] So, from first law of thermodynamics, i.e., \[\Delta Q=\Delta U+\Delta W\] We have \[1000=(300+x)+0\] i.e., \[x=1000-300=700\text{ }J\] As efficiency of a cycle of defined as \[\eta =\frac{work\,done}{input\,heat}\] \[=\frac{\Delta W}{{{Q}_{1}}+{{Q}_{4}}}=\frac{\Delta Q}{({{Q}_{1}}+{{Q}_{4}})}\] \[=\frac{1000}{(6000+3500)}=\frac{1000}{9500}\] \[=0.105\] \[=10.5%\]
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