Respuesta :
Explanation:
According to the heat relationship,
Heat given by steam = heat taken by water
Hence, [tex](mC_{p} \Delta T)_{steam}[/tex] = [tex](mC_{p} \Delta T)_{water}[/tex]
where, [tex]C_{p}[/tex] = specific heat
Value of [tex]C_{p}_{steam}[/tex] at 250^{o}C and 12.5 atm is 2.326 [tex]kJ/Kg^{o}C[/tex].
And, [tex]C_{p}_{steam}[/tex] at 200^{o}C and 12.5 atm is 2.743 [tex]kJ/Kg^{o}C[/tex].
Therefore, average [tex]C_{p}_{steam}[/tex] = [tex]\frac{2.326 kJ/Kg^{o}C + 2.743 kJ/Kg^{o}C}{2}[/tex]
= 2.5345 kJ/Kg^{o}C
As the given data is as follows.
[tex]m_{steam}[/tex] = 400 kg/hr, [tex]\Delta T_{steam}[/tex] = [tex]250^{o}C - 200^{o}C[/tex] = [tex]50^{o}C[/tex], [tex]m_{water}[/tex] = 500 kg/hr
[tex]C_{p}_{water}[/tex] = 4.186 [tex]kJ/kg^{o}C[/tex]
Hence, putting the given values into the above formula as follows.
[tex](mC_{p} \Delta T)_{steam}[/tex] = [tex](mC_{p} \Delta T)_{water}[/tex]
[tex](400 \times 2.5345 kJ/kg^{o}C \times 50^{o}C)_{steam}[/tex] = [tex](500 kg \times 4.186 kJ/kg^{o}C \Delta T)_{water}[/tex]
[tex]\Delta T_{water} = 24.2^{o}C[/tex]
As we know that, [tex]\Delta T_{water} = T_{2} - T_{1}[/tex]
[tex]T_{2} - 20^{o}C = 24.2^{o}C[/tex]
[tex]T_{2} = 44.2^{o}C[/tex]
Thus, we can conclude that the new temperature of water is [tex]44.2^{o}C[/tex].
