Q1 of 51 Page 62

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.


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4

Refer to figure. Let ΔU1 and ΔU2 be the change in internal energy in processes A and B respectively, ΔQ be the net heat given to the system in process A + B and ΔW be the net work done by the system in the process A + B.


5

The internal energy of an ideal gas decreases by the same amount as the work done by the system.

2

Figure shows a paddle wheel coupled to a mass of 12 kg through fixed frictionless pulleys. The paddle is immersed in a liquid of heat capacity 4200 JK–1 kept in an adiabatic container. Consider a time interval in which the 12 kg block falls slowly through 70 cm.

(a) How much heat is given to the liquid?


(b) How much work is done on the liquid?


(c) Calculate the rise in the temperature of the liquid neglecting the heat capacity of the container and the paddle.



3

A 100 kg block is started with a speed of 2.0 m
s–1 on a long, rough belt kept fixed in a horizontal position. The coefficient of kinetic friction between the block and the belt is 0.20.

(a) Calculate the change in the internal energy of the block-belt system as the block comes to a stop on the belt.


(b) Consider the situation com a frame of reference moving at 2.0 m s–1 along the initial velocity of the block. As seen from this frame, the block is gently put on a moving belt and in due time the block starts moving with the belt at 2.0 ms–1. Calculate the increase in the kinetic energy of the block as it stops slipping past the belt.


(c) Find the work done in this frame by the external force holding the belt.