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

Step-by-step 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 * 4.0177 kJ/kg^(o)C * (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`.