Question

a) Define entropy as a thermodynamic state variable. b) Determine the final temperature when a block of copper at 0 °C is added to 1.0 kg liquid water at 80 °C in an insulated container at constant atmospheric pressure. The copper has a heat capacity of 0.01 kJ kg1 and the specific heat of water is 4.2 kJ kg1 K (both assumed constant with temperature) c) Calculate the change in entropy of the system in b). s it appropriate to consider the 80 °C water as a thermal reservoir? d)

Answer 1

The concept used in this question is that ,When a hot body comes in contact with the cold body then the cold body absorbed heat , while the hot body released heat and finally both the bodies comes to the final temperature or the equilibrium temperature and then further no change in temperature takes place of both bodies , In order to prove entropy as a thermodynamic state variable just we have to prove then entropy change is equal to zero which is possible in case of the state function which depends only on the initial and final state of the system ,Entropy change formula is different for constant temperature body and the variable temperature body , in case of the variable temperature body we just apply integration as it is shown clearly in my answer..

a new stage 1 stager ---m y quer - stage consider a Reversible to which heat quer is isothermal fonoress supplied Entropy change from stage 1 to stage 2 As 1-2 = + Qoreu Entropy change DS2-1 = from stage a to stage a q nev

As total = Asing + As ay I tarev - Creu Tas total = 0 is is equal a state to hero which means entropy dependent property.

(6) het attained Te be the equilibrium Temperature By Liquid and copper Block (Heat losst by water) = (Heat gained by copper block) 4.2 X 10x ( 80-T) = 0.01x1 Ts-o) 4.2 X10 X 80 - 4.281.0 T = 0.01% TA T = 336.0 T = 79.000 Hence the final temperature on the equilibrium temperature is 49.boe. AS = (AS) watesn t (AS) copper block → change in entropy on the system (As) water = - 1.0 X 4.2 X 180-19.80) 180 + 273.15)

(DS) water - - 0.64 (KI) 353.15 (as)coperbank Spotting do = epdt ep Heat capacity for the copper block [ (15) copper = 1 (65) = ful 9) * (0.01 KE (AS) copper block = 0.01x ln ( 19.66 + 293.15 0+273.15 J 2 6-01 X 0.25630 = 0.002563 KI

AS = (-0.0093785 +0.002563) Ky As = 1.6458104 kJ IDS = 0.0001845 kJ Hence, the change in centropy of the system is 0.0001845 KI Yes, It is appropriate consider 80 l water as a thermal Reservoir Becau as we can see from the calculation that the temperature of water drop from 80°c to 79.60e, which means a minute Droß in temperature takes

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