Question

Water is heated in an insulated, constant-diameter tube by a \(7-\mathrm{kW}\) electric resistance heater. If the water enters the heater steadily at \(20^{\circ} \mathrm{C}\) and leaves at \(75^{\circ} \mathrm{C}\), determine the mass flow rate of water.

Short answer

Answer: The mass flow rate of water is approximately 0.0304 kg/s.

Step-by-step solution

Identify the given information and the formula to use

We are given the following information: - Power (P) = 7 kW - Initial Temperature (T1) = \(20^{\circ} \mathrm{C}\) - Final Temperature (T2) = \(75^{\circ} \mathrm{C}\) The specific heat capacity (c) of water is approximately \(4.18 \, \mathrm{kJ/kg \cdot K}\). We need to determine the mass flow rate (m') of water using the formula: \(P = m' \cdot c \cdot \Delta T\)

Calculate the temperature change

We need to find the temperature change (\(\Delta T\)) of the water as it passes through the heater: \(\Delta T = T2 - T1\) \(\Delta T = 75^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = 55 \, \mathrm{K}\)

Rearrange the formula to solve for the mass flow rate

We need to find the mass flow rate (m'), so let's rearrange the formula: \(m' = \dfrac{P}{c \cdot \Delta T}\)

Calculate the mass flow rate of water

Now, we can plug the values into the rearranged formula: \(m' = \dfrac{7 \, \mathrm{kW}}{4.18 \, \mathrm{kJ/kg \cdot K} \cdot 55 \, \mathrm{K}}\) \(m' = \dfrac{7 \, \mathrm{kW}}{229.9 \, \mathrm{kJ/kg}}\) \(m' \approx 0.0304 \, \mathrm{kg/s}\) The mass flow rate of water is approximately \(0.0304 \, \mathrm{kg/s}\).