A) Independent of heat given to the system
B) More than the heat given to system
C) Equal to heat given to the system
D) zero
Correct Answer: C
Solution :
Key Idea: In a cyclic process, system returns to its initial state. If and amount of heat Q is given to a system a part of it will be used in increasing the internal energy\[\left( Mk_{1}^{2} \right){{\omega }_{1}}=\left( Mk_{2}^{2} \right){{\omega }_{2}}\] of the system and the rest is doing work \[\Rightarrow \] by the system. Therefore, \[\frac{k_{1}^{2}}{k_{2}^{2}}=\frac{{{\omega }_{2}}}{{{\omega }_{1}}}\] (first law of thermodynamics) The internal energy of the system does not change in a cyclic process\[\Rightarrow \] because it is a function of only the state of the system. \[\frac{{{k}_{1}}}{{{k}_{2}}}=\frac{\sqrt{{{\omega }_{2}}}}{\sqrt{{{\omega }_{1}}}}\] \[\text{mg sin a}-{{f}_{\text{k}}}.\] Hence, in a cyclic process, the heat given to a system equal the net work done by the system.
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