If heat absorbed by the working substance from the source is $Q_{1} = 550 J$ , heat rejected to the sink is $Q_{2} = 200 J$ then work done by the working substance in a cyclic process is

250 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

350 J

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

750 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

550 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $350 J$
Heat absorbed by working substance,
$Q_{1} = + 550 J$
Heat rejected to the sink,
$Q_{2} = - 200 J$
For a cyclic process, change in internal energy $Δ U = 0$ , applying first law of thermodyamics
$\Rightarrow Q = Δ U + W . . . (i)$
For the cyclic process, $Q = Q_{1} + Q_{2}$
$\Rightarrow (550 + (- 200)) = 0 + W$
$∴ W = 350 J$

Suggest Corrections

Similar questions

Q. If heat absorbed by the working substance from the source is

Q_{1} = 550 J

, heat rejected to the sink is

Q_{2} = 200 J

then work done by the working substance in a cyclic process is

In a heat engine the ratio of net work done per cycle by the engine to the total amount of heat absorbed per cycle by the working substance from the source is known as

Q. In a Carnot's cycle, the working substance absorbs heat

Q_{1}

at temperature

T_{1}

and rejects heat

Q_{2}

at temperature

T_{2}

. The change of entropy during the Carnot's cycle is :

Q. If work done by the heat engine is

600 J

and heat absorbed by the working substance is

800 J

, the efficiency of the heat engine is

Q. If work done by the heat engine is

600 J

and heat absorbed by the working substance is

800 J

, the efficiency of the heat engine is

Join BYJU'S Learning Program

If heat absorbed by the working substance from the source is Q1=550 J, heat rejected to the sink is Q2=200 J then work done by the working substance in a cyclic process is

The correct option is B 350 JHeat absorbed by working substance, Q1=+550 J Heat rejected to the sink, Q2=−200 J For a cyclic process, change in internal energy ΔU=0, applying first law of thermodyamics ⇒Q=ΔU+W ...(i) For the cyclic process, Q=Q1+Q2 ⇒(550+(−200))=0+W ∴W=350 J

If heat absorbed by the working substance from the source is $Q_{1} = 550 J$ , heat rejected to the sink is $Q_{2} = 200 J$ then work done by the working substance in a cyclic process is