Question

In a shower, cold water at \(10^{\circ} \mathrm{C}\) flowing at a rate of \(5 \mathrm{kg} / \mathrm{min}\) is mixed with hot water at \(60^{\circ} \mathrm{C}\) flowing at a rate of \(2 \mathrm{kg} / \mathrm{min} .\) The exit temperature of the mixture is \((a) 24.3^{\circ} \mathrm{C}\) (b) \(35.0^{\circ} \mathrm{C}\) \((c) 40.0^{\circ} \mathrm{C}\) \((d) 44.3^{\circ} \mathrm{C}\) \((e) 55.2^{\circ} \mathrm{C}\)

Short answer

Answer: (a) 24.3°C

Step-by-step solution

Identify the mass flow rates of the waters

The cold water is flowing at a rate of 5 kg/min and the hot water is flowing at a rate of 2 kg/min.

Calculate the total mass flow rate of the mixture

To find the total mass flow rate of the mixture, we add the mass flow rate of cold water to the mass flow rate of hot water. Total mass flow rate = Mass flow rate of cold water + Mass flow rate of hot water = 5 kg/min + 2 kg/min = 7 kg/min

Calculate the heat gained by the cold water

Heat gained by cold water (Q_cold) can be found using the formula: Q_cold = mass flow rate of cold water × specific heat of water × (Exit temperature - Initial temperature of cold water) Q_cold = 5 kg/min × 4.18 kJ/kg°C × (Exit temperature - 10°C)

Calculate the heat lost by the hot water

Heat lost by hot water (Q_hot) can be found using the formula: Q_hot = mass flow rate of hot water × specific heat of water × (Initial temperature of hot water - Exit temperature) Q_hot = 2 kg/min × 4.18 kJ/kg°C × (60°C - Exit temperature)

Set up an equation to relate heat gained and heat lost

Since the heat gained by the cold water is equal to the heat lost by the hot water, we can set up an equation: Q_cold = Q_hot 5 kg/min × 4.18 kJ/kg°C × (Exit temperature - 10°C) = 2 kg/min × 4.18 kJ/kg°C × (60°C - Exit temperature)

Solve the equation to find the Exit temperature

Now, we need to solve the equation for the Exit temperature: 5 × (Exit temperature - 10°C) = 2 × (60°C - Exit temperature) Expanding, we get: 5 × Exit temperature - 50°C = 120°C - 2 × Exit temperature Adding 2 × Exit temperature to both sides: 7 × Exit temperature = 170°C Divide both sides by 7: Exit temperature = 170°C / 7 Exit temperature = 24.29°C By comparing the given options, the exit temperature is closest to \((a) 24.3^{\circ} \mathrm{C}.\)