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Ghella [55]
3 years ago
5

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).
Physics
1 answer:
Tamiku [17]3 years ago
4 0

Explanation:

The given data is as follows.

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

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

     T_{2} = T - 25

Now, according to the law of conservation of energy,

         m_{1}C_{1}T_{1} = m_{2}C_{2}T_{2} + mL

     800 \times 4.18 \times (75 - T) = 1000 \times 4.18 \times (T - 25) + 30000

      250.8 - 3.34T = 4.18T - 104.5 + 30

          T = \frac{325.3}{7.52}

              = 43.23^{o}C

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

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