A thermally insulated, closed copper vessel contains water at 15°C. When the vessel is shaken vigorously for 15 minutes, the temperature rises to 17°C. The mass of the vessel is 100 g and that of the water is 200 g. The specific heat capacities of copper and water are 420 J kg–1 K–1 and 4200 J kg–1 K–1 respectively.
Neglect any thermal expansion.
(a) How much heat is transferred to the liquid-vessel system?
(b) How much work has been done on this system?
(c) How much is the increase in internal energy of the system?
Given
Initial temperature of water T1=15℃ =288K
Final temperature of water T2=17℃ =290K
Specific heat capacity of copper cc=420 J kg–1 K–1
Specific heat capacity of water cw=4200 J kg–1 K–1
Mass of copper vessel mc=100g = 100×10-3kg
Mass of water mw=200g = 200×10-3kg
a) It is given that copper vessel is thermally insulated. Therefore, no heat from the surroundings can be transferred to the liquid-vessel system.
b) Work done on this system will be
ΔW=mwcwΔT + mcccΔT
ΔT=T2-T1=290-288=2K
So,
ΔW= 200×10-3×4200×2 + 100×10-3×420×2
ΔW=1680 + 84=1764J
∴ work done on the system is 1764J.
c) 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
From part (a) we have concluded that ΔQ=0
So, first law becomes
ΔU=-ΔW
Work done on the system ΔW=1764 J
So, work done by the system ΔW=-1764J
ΔU = -(-1764) = 1764J
∴ Increase in internal energy of the system = 1764J.