Q. 125.0( 1 Vote )

# A substance is taken through the process abc as shown in the figure. If the internal energy of the substance increases by 5000 J and heat of 2625 cal is given to the system, calculate the value of J.

‘J' is mechanical equivalent of heat a conversion factor between two different units of energy: calorie to the joule.

Given

Heat given to system =2625cal =2625×J J

Change in internal energy= 5000J

From graph

Va=0.02m3

Vb=Vc=0.05m3

Pa=Pb=200kPa=200×103Pa

Pc=300kPa=300×103Pa

We know that work done by the gas is given as

ΔW=PΔV

Work done in process abc=ΔW=Wab+Wbc

ΔW=Pa(Vb-Va)+0 ( Wbc=0 because Vb=Vc)

ΔW=200×103×(0.05-0.02)=6000J

From first law of thermodynamics, we know that,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

ΔQ=5000+6000=11000J

But ΔQ=2625cal =2625×J J

Therefore,

2625×J=11000

value of ‘J’ is 4.19joule/cal.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
JEE Advanced Level Questions of NLM48 mins
Work done in Thermodynamic ProcessFREE Class
Microsporogenesis & Megasporogenesis49 mins
Meselson and Stahl experiment49 mins
NEET 2021 | Transcription - An important Topic62 mins
DNA Fingerprinting42 mins
Interactive Quiz on Sexual Reproduction in flowering plants44 mins
Pedigree chart58 mins
MCQs of Ecology for NEET52 mins
Different high order thinking questions based on DNA structure39 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses