Please I need a clearly written, well solved and explanatory solution. I have ha
ID: 1566572 • Letter: P
Question
Please I need a clearly written, well solved and explanatory solution. I have had misleading solutions in the past which was really annoying. I appreciate a question left unsolved than a misleading solution. Please do not attempt if you are not sure that you can solve it. Thanks
An insulated container of volume 0.160 m is divided into two equal chambers of volume 0.0800 ms. Chamber A is completely filled with water at a temperature of 20 °C and a pressure of 100 kPa. Chamber B contains a copper block at a temperature of 100 oC. The insulating partition between A and Bis then removed and the water and copper block interchange heat until a thermal equilibrium is reached. i. What is the final temperature of the water? i. What is the final temperature of the copper block? ii. How much heat energy was transferred from the copper block to the water? iv. What is the resulting change in the entropy after the partition is removed?Explanation / Answer
PART-1 and 2: The maximum temperature mixture can attain is 100oC and the lowest temperature it can attain is 20oC. So the final temperature of the system is between the two temperatures.
Heat absorbed (positive) or released (negative) by a system is given as, Q = + m*C*(tf-ti)
Here 'm' is mass of the system, 'C' is specific heat capacity, 'ti' and 'tf' are initial and final temperature of the system.
Now, heat taken / absorbed = heat given / released
Chamber A is at low temperature so it will absorb heat and chamber B is at high temperature so it will release heat. The final temperature of both the chambers will be same.
So the above expression can be written as,
mA*C*( tf - tiA ) = - mB*C*( tf - tiB)
(VADA)*C*( tf - tiA ) = - (VBDB)*C*( tf - tiB) DA= 1000kg/m3, DB= 8890kg/m3 (constant)
0.0800m3*1000kg/m3*C*( tf - 20oC ) = 0.0800m3*8890kg/m3*C*( tf - 100oC )
From here we get the value of final temperature when the system is in euilibrium,
tf = 91.9oC (final temperature of water and copper block)
PART-3: Heat transferred from the copper block to water is given by,
Q = mB*C*( tf - tiB)
Q = (VBDB)*C*( tf - tiB) ( specific heat of copper = 390J/kgoC )
Q = (0.0800m3*8890kg/m3)*(390J/kgoC)*( 91.9oC - 100oC )
Q = - 2.246 * 106 J (negative sign because heat is released from the copper block)
PART-4: Resulting change in entropy is given by,
dQ = TdS
dS = dQ / T here, dQ heat given or taken between A and B, dS is change in entropy of the system and T is the temperature.
dS = ( - Q/ TB ) + ( Q/ TA )
dS = ( - 2.246 * 106 J/ 100oC ) + ( 2.246 * 106J / 20oC )
dS = 8.98 * 104 J/oC