Q20 of 51 Page 62

Calculate the increase in the internal energy of 10g of water when it is heated from 0°C to 100°C and converted into steam at 100 kPa. The density of steam = 0.6 kg m–3. The specific heat capacity of water = 4200 J kg–1 °C–1 and the latent heat of vaporization of water = 2.25 × 106 J kg–1.

Given


The density of steam ρ’= 0.6 kg m–3


Mass of water m=10g =0.010kg


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


latent heat of vaporization of water L = 2.25 × 106 J kg–1.


Pressure P =100kPa =100×105Pa


Change in temperature ΔT= (100-0) oC =100oC


Density of water ρ =1000 kg m-3


We know that specific heat capacity is given by



Where ΔQ = heat supplied to the system


Therefore, ΔQ= cmΔT


Also, ΔQ =mL


Where m= mass of the substance


L=latent heat


Therefore, ΔQ=mL + cmΔT


=0.010×2.25 × 106 + 4200×0.01×100


=22500+4200


=26700J


We know that work done by the gas is given as


ΔW=PΔV


Where ΔV =change in volume


P =pressure


Also,





ΔW=105×0.01699=1699 J


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


ΔU=ΔQ-ΔW


=26700-1699


=25001J


Thus, change in internal energy is 25001J.


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Find the change in the internal energy of 2 kg of water as it is heated from 0°C to 4°C. The specific heat capacity of water is 4200 J kg–1 K–1 and its densities at 0°C and 4°C are 999.9 kg m–3 and 1000 kg m–3 respectively. Atmospheric pressure = 105 Pa.

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Figure shows a cylindrical tube of volume V with adiabatic walls containing an ideal gas. The internal energy of this ideal gas is given by 1.5 nRT. The tube is divided into two equal parts by a fixed diathermic wall. Initially, the pressure and the temperature are p1, T1 on the left and p2, T2 on the right. The system is left for sufficient time so that the temperature becomes equal on the two sides.

(a) How much work has been done by the gas on the left part?


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(d) How much heat has flown from the gas on the right to the gas on the left?



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An adiabatic vessel of total volume V is divided into two equal parts by a conducting separator. The separator is fixed in this position. The part on the left contains one mole of an ideal gas (U = 1.5 nRT) and the part on the right contains two moles of the same gas. Initially, the pressure on each side is p. The system is left for sufficient time so that a steady state is reached. Find

(a) the work done by the gas in the left part during the process.


(b) the temperature on the two sides in the beginning,


(c) the final common temperature reached by the gases,


(d) the heat given to the gas in the right part and


(e) the increase in the internal energy of the gas in the left part.