Question

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).

Answer 1

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 $43.23^{o}C$.