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A stream of warm water is produced in a steady-flow mixing process by combining 1.0 kg/s of cool water at 25 °C with 0.8 kg/s of hot water at 75 °C. During mixing, heat is lost to the surroundings at the rate of 30 kJ/s. What is the temperature of the warm water stream? Assume the specific heat of water is constant at 4.18 kJ/(kg·K).

Respuesta :

Explanation:

The given data is as follows.

      [tex]m_{1}[/tex] = 0.8 kg/s = 800 g,      [tex]C_{1} = 4.18 J/g^{o}C[/tex]

     [tex]m_{2}[/tex] = 1 kg = 1000 g,   [tex]T_{1}[/tex] = (75 - T),

     [tex]T_{2}[/tex] = T - 25

Now, according to the law of conservation of energy,

         [tex]m_{1}C_{1}T_{1} = m_{2}C_{2}T_{2} + mL[/tex]

     [tex]800 \times 4.18 \times (75 - T) = 1000 \times 4.18 \times (T - 25)[/tex] + 30000

      250.8 - 3.34T = 4.18T - 104.5 + 30

          T = [tex]\frac{325.3}{7.52}[/tex]

              = [tex]43.23^{o}C[/tex]

Thus, we can conclude that temperature of the warm water stream is [tex]43.23^{o}C[/tex].

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