An air handler is a large unmixed heat exchanger used for comfort control in large buildings. In one such application, chilled water \(\left(c_{p}=4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters an air handler at \(5^{\circ} \mathrm{C}\) and leaves at \(12^{\circ} \mathrm{C}\) with a flow rate of \(1000 \mathrm{~kg} / \mathrm{h}\). This cold water cools \(5000 \mathrm{~kg} / \mathrm{h}\) of air \(\left(c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) which enters the air handler at \(25^{\circ} \mathrm{C}\). If these streams are in counter-flow and the water-stream conditions remain fixed, the minimum temperature at the air outlet is (a) \(5^{\circ} \mathrm{C}\) (b) \(12^{\circ} \mathrm{C}\) (c) \(19^{\circ} \mathrm{C}\) (d) \(22^{\circ} \mathrm{C}\) (e) \(25^{\circ} \mathrm{C}\)

The energy balance equation states that the energy entering or leaving a system must be equal to the energy entering or leaving it. For this problem, we can apply an energy balance on the air side, assuming no additional energy is exchanged with the surroundings: \(\dot{m}_{air} c_{p, air} (T_{in, air} - T_{out, air}) = \dot{m}_{water} c_{p, water} (T_{in, water} - T_{out, water})\) where \(\dot{m}_{air}\) and \(\dot{m}_{water}\) are the mass flow rates of the air and water respectively; \(c_{p, air}\) and \(c_{p, water}\) are the specific heat capacities of air and water, respectively; and \(T_{in, air}\), \(T_{out, air}\), \(T_{in, water}\), and \(T_{out, water}\) are the inlet and outlet temperatures of the air and water, respectively.

The mass flow rates of air and water are given, as well as their respective specific heat capacities and inlet and outlet temperatures. Using these values and plugging them into the energy balance equation, we have: \((5000 \mathrm{~kg} / \mathrm{h}) (1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(25^{\circ} \mathrm{C} - T_{out, air}) = (1000 \mathrm{~kg} / \mathrm{h}) (4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(5^{\circ} \mathrm{C} - 12^{\circ} \mathrm{C})\) Now we can solve for the unknown temperature \(T_{out, air}\): \(5000(25 - T_{out, air}) = -1000(4.2)(5 - 12)\) \(125000 - 5000 T_{out, air} = 29400\) \(-5000 T_{out, air} = -95500\) \(T_{out, air} = 19.1^{\circ} \mathrm{C}\) Since the air is cooled in a heat exchanger, it is not possible to obtain a lower temperature at the air outlet because the water temperature is fixed. Therefore, the minimum temperature at the air outlet is (c) \(19^{\circ} \mathrm{C}\).