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Question 1.134.E: In order to cool 1 short ton (2000 lbm) of water at 70°F in ...

In order to cool 1 short ton (2000 lbm) of water at 70°F in a tank, a person pours 160 lbm of ice at 25°F into the water. Determine the final equilibrium temperature in the tank. The melting temperature and the heat of fusion of ice at atmospheric pressure are 32°F and 143.5 Btu/lbm, respectively.

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A 1-short ton (2000 lbm) of water at 70°F is to be cooled in a tank by pouring 160 lbm of ice at 25°F into it. The final equilibrium temperature in the tank is to be determined. The melting temperature and the heat of fusion of ice at atmospheric pressure are 32°F and 143.5 Btu/lbm, respectively

Assumptions 1 Thermal properties of the ice and water are constant. 2 Heat transfer to the water is negligible.

Properties The density of water is 62.4  lbm / ft ^{3}, and the specific heat of water at room temperature is C =  1.0  Btu / lbm \cdot{ }^{\circ} F (Table A-9). The heat of fusion of ice at atmospheric pressure is 143.5 Btu/lbm and the specific heat of ice is 0.5  Btu / lbm .{ }^{\circ} F.

Analysis We take the ice and the water as our system, and disregard any heat transfer between the system and the surroundings. Then the energy balance for this process can be written as

\underbrace{E_{i n}-E_{\text {out }}}_{\begin{array}{c} \text { Net energy transfer } \\\text { by heat, work, and mass }\end{array}}=\underbrace{\Delta E_{\text {system }}}_{\begin{array}{c}\text { Change in internal, kinetic, } \\\text { potential, etc. energies }\end{array}}

 

                        \rightarrow 0=\Delta U \rightarrow(\Delta U)_{\text {ice }}+(\Delta U)_{\text {water }}=0

 

\left\lfloor m C\left(32^{\circ} F -T_{1}\right)_{\text {solid }}+m h_{i f}+m C\left(T_{2}-32^{\circ} F \right)_{\text {liquid }}\right\rfloor_{\text {ice }}+\left[m C\left(T_{2}-T_{1}\right)\right]_{\text {water }}=0

 

Substituting,

(160  lbm )\left[\left(0.50  Btu / lbm \cdot{ }^{\circ} F \right)(32-25)^{\circ} F +143.5  Btu / lbm +\left(1.0  Btu / lbm \cdot{ }^{\circ} F \right)\left(T_{2}-32\right){ }^{\circ} F \right]

 

                                   + (2000  lbm )\left(1.0  Btu / lbm \cdot{ }^{\circ} F \right)\left(T_{2}-70\right){ }^{\circ} F =0

 

It gives              T_{2}= 5 6 . 3 ^{\circ} F

 

which is the final equilibrium temperature in the tank.

112

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