Respuesta :
Explanation:
The given data is as follows.
[tex]T_{1} = 83^{o}C[/tex] = (83 + 273) K = 356 K
[tex]T_{2} = 100^{o}C[/tex] = (100 + 273) K = 373 K
m = 3.9 kg/s
Relation between heat energy and specific heat is as follows.
Q = [tex]mC_{p} \Delta T[/tex]
Putting the given values into the above formula as follows.
Q = [tex]mC_{p} \Delta T[/tex]
= [tex]3.9 kg/s \times 4200 J/kg.K \times (373 - 356)K[/tex]
= 278460 J/s
As it is given that enthalpy of evaporation of water is 2760000 J/kg.
Hence, energy left is as follows.
(2760000 - 278460) J/s
= 2481540 J/s
So, enthalpy of vapor energy at [tex]100^{o}C[/tex] is as follows.
[tex]3.9 kg/s \times 2760000 J/kg[/tex]
= 10764000 J/s
Hence, energy left for steam is as follows.
(10764000 J/s - 2481540 J/s)
= 8282460 J/s
Now, steam is formed at [tex]100^{o}C[/tex] or (100 + 273)K = 373 K. Therefore, final temperature will be calculated as follows.
Q = [tex]mC_{p} \Delta T[/tex]
8282460 J/s = [tex]3.9 kg/s \times 1800 J/kg.K \times (T_{2} - 373)K[/tex]
[tex]T_{2}[/tex] = 1552.83 K
Thus, we can conclude that the temperature of steam leaving the pipe is 1552.83 K.
