8 8) Steam at 232.1 kPa is used to heat a food product with a specific heatof 4.05 kJ/(kg K). The food is heated from 20 to 70 °C. 3 kg steam isused as saturated vapor. Steam exits as saturated liquid at 95 °C.a) What is the initial temperature of the steam?b) Choose a basisc) How much food product is heated?d) How much food product would have been heated if the steam enters assaturated liquid at 232.1 kPa and leaves as saturated liquid at 95 °℃?

(a) To find: The initial temperature of the steam. Given: The steam pressure is P=232.1 kPa. The specific heat is cf=4.05 kJkg·K. The initial temperature of food is Ti=20°C and final temperature of food is Tf=70°C. The mass of steam is m=3 kg. The steam exits as saturated liquid at T3=95°C. TV diagram: The TV diagram shown below:Calculation: The specific enthalpy and saturation temperature at 1 corresponding to P=232.1 kPa and x1=1 is, Tsat=125°Ch1=2713.1 kJkg Thus, the initial temperature of steam is 125°C.(b) To choose: The basis. Explanation: Convection: The heat transfer to the body by movement of heat fluid is termed as convection. The basis of food heating is convection.(c) To find: The amount of food product is heated. Formula used: The heat rejected by steam is heat observed by the food. mh1-h3=mfcfTf-Ti Here, h3 is enthalpy of steam at exit and mf is mass of food. Calculation: The enthalpy at 3 corresponding to T3=95°C is, h3=hf=398.09 kJkg Substitute respective values in above equation. 32713.1-398.09=mf4.0570-2034.43 kg=mf Thus, the amount of food product is heated is 34.43 kg.(d) To find: The amount of food product is heated is steam enters at saturated liquid and leaves at saturated liquid. Formula used: The heat rejected by steam is heat observed by the food. mh2-h3=mfcfTf-Ti Here, h3 is enthalpy of steam at exit and mf is mass of food. Calculation: The enthalpy at 2 corresponding to P=232.1 kPa is, h2=hf=525.07 kJkg Substitute respective values in above equation. 3525.0-398.09=mf4.0570-201.88 kg=mf Thus, the amount of food product is heated is 1.88 kg.