Answer :

Given


Mass of water m = 2kg


Change in temperature T =4 -0 =4


Specific heat capacity of water c =4200 J kg–1 K–1


Density of water at 0°C=999.9 kg m–3


Density of water at 4°C=1000 kg m–3


Atmospheric pressure P = 105 Pa.


We know that specific heat capacity is given by



Where ΔQ = heat supplied to the system


Therefore, ΔQ=cmΔT


=4200×2×4=33600J


We know that work done by the gas is given as


ΔW=PΔV


Where ΔV =change in volume


P =pressure


Also,



Volume at 0oC V1



Similarly, volume at 4oC V2



ΔW=P(V2-V1)



From first law of thermodynamics, we know that,


Q=U+ΔW


Where ΔQ=heat supplied to the system


ΔU=change in internal energy


ΔW=work done by the system


ΔU=ΔQ-ΔW


=33600-(-0.02)


=33599.98J


Thus, change in internal energy is 33599.98J.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
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
caricature
view all courses
RELATED QUESTIONS :

<span lang="EN-USHC Verma - Concepts of Physics Part 2

<span lang="EN-USHC Verma - Concepts of Physics Part 2

<span lang="EN-USHC Verma - Concepts of Physics Part 2

<span lang="EN-USHC Verma - Concepts of Physics Part 2

A thermally insulHC Verma - Concepts of Physics Part 2

<span lang="EN-USHC Verma - Concepts of Physics Part 2

<span lang="EN-USHC Verma - Concepts of Physics Part 2

50 cal of hHC Verma - Concepts of Physics Part 2

<span lang="EN-USHC Verma - Concepts of Physics Part 2

Consider two procHC Verma - Concepts of Physics Part 2