Citrus Concentrate

Great orange juice is made from fresh oranges sliced and squeezed. The alternative is to prepare the juice from a concentrated product. Water and valuable oils are removed from fresh citrus at a processing plant then "concentrate" is shipped to consumers with the instructions,

"Mix concentrate with 3 cans of tap water."

Suppose the precise initial temperature of the solid concentrate and tap water is known:

Basic Thermodynamics ~ J. Pohl ©

www.THERMOspokenhere.com (171-D130)

Citrus Concentrate

Great orange juice is made from fresh oranges sliced and squeezed. The alternative is to prepare the juice from a concentrated product. Water and valuable oils are removed from fresh citrus at a processing plant then "concentrate" is shipped to consumers with the instructions,

"Mix concentrate with 3 cans of tap water."

Suppose the precise initial temperature of the solid concentrate and tap water is known:

1. Can the second temperature, the temperature of the mixed OJ be determined
2. Can any precise, second LOWEST temperature of the OJ be determined? Yes! If the mixing goes very slowly the OJ will be served at room temperature."
3. OJ is best served cold. How cold might it be? If we assume no heat with the surroundings - that second temperature will be the least possible BUT not attained."

Calculate the very least possible temperature of OJ mixed.
♦  The concentrate is "one volume" of solid water @ - 15°C.

Specific Heat Solid cp,solid,avg = 1.36 J/(g°C)

Heat of Fusion hsf = -333.4 J/g

Specific Heat Liquid cp,liquid,avg = 4.16 J/(g°C)

The three volumes of water are initially 20°C.

Solution: Mass Equation: The initial state of orange juice is a mix of "one can" of concentrate, with "3 cans" of tap water.

The mass equation states simply that the mass of the system finally will be the mass of the concentrate and water

.

mass_{sys, initial}= mass_{(solid water, -15°C)}+ mass_{(liquid water, 20°C)}

mass_{sys, final} = mass_{[(all solid at ?C) or (all liquid at ?C) or (solid and liquid at 0°C]}

So the Mass Equation for this mixing of water event is:

m₂  -  m₁ = 0

[mass_{[(all solid at ?C) or (all liquid at ?C) or (solid and liquid at 0°C]}] - [mass_{(solid water, -15°C)} + mass_{(liquid water, 20°C)}] = 0

One might ponder how large is the can, does the can size matter? The mass of a "can" of solid concentrate is essentially the same as the mass of that "can" filled with water. (The density of water as ice or water is about 4% less than as liquid).

The mass here is a mass of "one can" of concentrate However, we are not mixing by mass, we are mixing by volumes. No matter. For this event the difference in liquid water mass per volume from solid mass per volume is insignificant.

The is event is called a thermal equilibration. Occurring at one atmosphere, the concentrate, initially ice at -15°C are placed together and attain the same temperature. We will assume the second state is liquid and solid at 0°C.

To write the energy equation we will identify three masses. The mass of ice as "ice_melt" which melts. Then "ice_solid" being that ice at -15°C that is warmed to 0°C but that remains solid and the water that is liquid at room temperature and is cooled to 0°C.

The mass equation is:

m₂ - m₁ = 0

But

          

m₂ = m_water,1 + m_ice,melt + m_ice,solid

And

m₁ = m_ice,1 + m_water,1

This Example has not been completed.