Hot oil \(\left(c_{p}=2.1 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(110^{\circ} \mathrm{C}\) and \(8 \mathrm{~kg} / \mathrm{s}\) is to be cooled in a heat exchanger by cold water \(\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) entering at \(10^{\circ} \mathrm{C}\) and at a rate of \(2 \mathrm{~kg} / \mathrm{s}\). The lowest temperature that oil can be cooled in this heat exchanger is (a) \(10.0^{\circ} \mathrm{C}\) (b) \(33.5^{\circ} \mathrm{C}\) (c) \(46.1^{\circ} \mathrm{C}\) (d) \(60.2^{\circ} \mathrm{C}\) (e) \(71.4^{\circ} \mathrm{C}\)

To calculate the heat transfer rate of the oil, we need to determine the mass flow rate and its specific heat. We are given that the mass flow rate of the hot oil is 8 kg/s and its specific heat is 2.1 kJ/(kg⋅K). Heat transfer rate: \(Q = m \times c_p \times \Delta T\) The temperature change of the oil, \(\Delta T\), is calculated as the initial temperature minus the final temperature: \(\Delta T = T_{initial} - T_{final}\) Since we are only interested in the minimum final temperature of the oil, we can determine the heat transfer rate as: \(Q = (8 \, \mathrm{kg/s}) \times (2.1 \, \mathrm{kJ/(kg \cdot K)}) \times (110 - T_{final})\)

To calculate the heat gain rate of the water, we need to determine the mass flow rate and its specific heat. We are given that the mass flow rate of the cold water is 2 kg/s, and its specific heat is 4.18 kJ/(kg⋅K). Heat gain rate: \(Q = m \times c_p \times \Delta T\) To calculate the temperature change of the water stream, we need to find the difference between its final temperature and the initial temperature. It is assumed that the minimum final temperature of the oil will be equal to the final temperature of the water: \(\Delta T = T_{final} - 10\) So we can determine the heat gain rate as: \(Q = (2 \, \mathrm{kg/s}) \times (4.18 \, \mathrm{kJ/(kg \cdot K)}) \times (T_{final} - 10)\)

To find the minimum final temperature of the oil, we need to set the heat transfer rate of the oil equal to the heat gain rate of the water: \((8 \times 2.1 \times (110 - T_{final})) = (2 \times 4.18 \times (T_{final} - 10))\)

Now, we need to solve the equation for the final temperature, \(T_{final}\): \(16.8 (110 - T_{final}) = 8.36 (T_{final} - 10)\) Expanding both sides, \(1848 - 16.8 T_{final} = 8.36 T_{final} - 83.6\) Adding \(16.8 T_{final}\) and \(83.6\) to both sides, we get: \(1931.6 = 25.16 T_{final}\) Dividing by 25.16 to find \(T_{final}\): \(T_{final} = \dfrac{1931.6}{25.16} = 76.72 \, \mathrm{C}\) Since the final temperature of the oil cannot be higher than the initial temperature of the water, we must subtract the initial temperature of the water (10°C): \(T_{final} = 76.72 - 10 = 66.72 \, \mathrm{C}\) However, this result is not present in the given options. The closest value is: \(T_{final} \approx 71.4 \, \mathrm{C}\) (e) Though the calculated value does not match exactly with an option, we can conclude that the lowest temperature that the oil can be cooled in this heat exchanger is approximately \(71.4^{\circ} \mathrm{C}\).