Question

A mixture of 454 kg of applesauce at 10 degrees Celsius is heated in a heat exchanger by adding 121300 kJ. Calculate the outlet temperature of the applesauce. (Hint: Heat capacity for applesauce is given at 32.8 degrees Celsius. Assume that this is constant and use this as the average.)

Answer 1

Explanation:

The given data is as follows.

           Mass of apple sauce mixture = 454 kg

           Heat added (Q) = 121300 kJ

 Heat capacity ($C_{p}$) of apple sauce at $32.8^{o}C$ = 4.0177 $kJ/kg^{o}C$

So, Heat given by heat exchanger = heat taken by apple sauce

                            Q = $mC_{p} \Delta T$

or,                    Q = $mC_{p} (T_{f} - T_{i})$  

Putting the given values into the above formula as follows.

                     Q = $mC_{p} (T_{f} - T_{i})$  

              121300 kJ = $454 kg \times 4.0177 kJ/kg^{o}C \times (T_{f} - 10)$

                      $T_{f}$ = $76.5^{o}C$

Thus, we can conclude that outlet temperature of the apple sauce is $76.5^{o}C$.