Answer:74.75 kJ/kg
Step-by-step explanation:
Given
mass of hot fluid
=5 kg/s
Energy associated with it=150 kJ/kg
mass of cold fluid
=15 kg/s
Energy associated with it=50 kJ/kg
Energy lost=5.5 kW
Let the Enthalpy of outlet fluid be h
mass at outlet
![m=m_l+m_h=20 kg/s](https://img.qammunity.org/2020/formulas/engineering/college/6tqteokj32bp9359qxld4qya9q5vy3xs3i.png)
Using Energy Conservation
![m_h* E_h+m_l* E_l+Q=m* h](https://img.qammunity.org/2020/formulas/engineering/college/vg7xs1davlaiz2wcq509n50zjym9hxwfjw.png)
![5* 150+15* 50-5.5=20* h](https://img.qammunity.org/2020/formulas/engineering/college/puznpdzyztfsza1qab4x5np0ctm4t11cbr.png)
h=74.75 kJ/kg