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Questions  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Thermodynamics

A heat engine absorbs heat q1 from a source at temperature $T_1$ and heat $ q_2$ from a source a temperature $T_2$. Work done is found to be $J(q_1+q_2)$. This is in accordance with

$\begin {array} {1 1} (A)\;\text{1st law of thermodynamics } & \quad (B)\;\text{2nd law of thermodynamics} \\ (C)\;\text{Joules equivalent law} & \quad (D)\;\text{None of these} \end {array}$

1 Answer

Joules law suggests that
J = Mechanical work done by the system, W/Net heat given to the system, Q
hence, $J =\large\frac{ W}{(q_1+q_2)}$
Hence, $W = J(q_1+q_2)$ is constant with the Joules law of equivalence
Ans : (C)
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