A stream of warm water is produced in a steady-flow mixing process by combining of cool water at 25°C with of hot water at 75°C. During mixing, heat is lost to the surroundings at the rate of 30 What is the temperature of the warm water stream? Assume the specific heat of water is constant at

Question

What will be an ideal response?

Answer 1

Here it may be helpful to draw a very simple schematic:



The above diagram assumes that you can do the mass balance ‘in your head’: 1 kg/s in plus 0.8 kg/s in = 1.8 kg/s out at steady state. We can write the enthalpy balance as

?H = Q? + ?

We are given that the heat removal rate is –30 kJ/s and that no work is done. So, we simply have

?H = ?30 kJ/s

The enthalpy change is the sum of the mass flow rates of the input streams each multiplied by their respective temperature change and by the heat capacity of water:

?H = 1 kg/s ? ( - 25). 4.18 kJ/(kg K) + 0.8 kg/s ?( – 75) ? 4.18 kJ/(kg K) = ?30 kJ/s

?H = 7.524 – 355.3 = ?30

from which = 43.2 °C

Note that because only temperature differences were involved in the problem (and not absolute temperatures) using °C for the units of temperature worked fine. A temperature difference is the same in °C as in K.