Q. 19

# <span lang="EN-US

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.

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