Respuesta :
Answer:
The value of total entropy change during the process
[tex]dS = 0.608 \frac{KJ}{K}[/tex]
Explanation:
mass of iron [tex]m_{iron}[/tex] = 25 kg
Initial temperature of iron [tex]T_{1}[/tex] = 350°c = 623 K
Mass of water [tex]m_{w}[/tex] = 100 kg
Initial temperature of water [tex]T_{2}[/tex] = 180°c = 453 K
When iron block is quenched inside the water the final temperature of both iron & water becomes equal. this is = [tex]T_{f}[/tex]
Thus heat lost by the iron block = heat gain by the water
⇒ [tex]m_{iron}[/tex] [tex]C_{iron}[/tex] ( [tex]T_{1}[/tex] - [tex]T_{f}[/tex] ) = [tex]m_{w}[/tex] [tex]C_{w}[/tex] ( [tex]T_{f}[/tex] - [tex]T_{2}[/tex] )
⇒ 25 × 0.448 × ( [tex]T_{1}[/tex] - [tex]T_{f}[/tex] ) = 100 × 4.2 × ( [tex]T_{f}[/tex] - [tex]T_{2}[/tex] )
⇒ [tex]( T_{1} - T_{f} ) = 37.5 ( T_{f} - T_{2} )[/tex]
⇒ [tex]( 623 - T_{f} ) = 37.5 ( T_{f} - 453 )[/tex]
⇒ [tex]( 623 - T_{f} ) = 37.5 T_{f} - 16987.5[/tex]
⇒ [tex]38.5 T_{f} = 17610.5[/tex]
⇒ [tex]T_{f} = 457.41 K[/tex]
This is the final temperature after quenching.
The total entropy change is given by,
[tex]dS = m_{iron}\ C_{iron} \ ln \frac{T_{f} }{T_{1} } + m_{w}\ C_{w} \ ln \frac{T_{f} }{T_{2} }[/tex]
Put all the values in above formula,
[tex]dS =[/tex] 25 × 0.448 × [tex]ln \frac{457.41}{623}[/tex] + 100 × 4.2 × [tex]ln \frac {457.41}{453}[/tex]
[tex]dS =[/tex] - 3.46 + 4.06
[tex]dS = 0.608 \frac{KJ}{K}[/tex]
This is the value of total entropy change.
