Explanation:
Formula for steady flow energy equation for the flow of fluid is as follows.
[tex]m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w[/tex]
Now, we will substitute 0 for both [tex]z_{1}[/tex] and [tex]z_{2}[/tex], 0 for w, 334.9 kJ/kg for [tex]h_{1}[/tex], 2726.5 kJ/kg for [tex]h_{2}[/tex], 5 m/s for [tex]V_{1}[/tex] and 220 m/s for [tex]V_{2}[/tex].
Putting the given values into the above formula as follows.
[tex]m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w[/tex]
[tex]1 \times [334.9 \times 10^{3} J/kg + \frac{(5 m/s)^{2}}{2} + 0] + q = 1 \times [2726.5 \times 10^{3} + \frac{(220 m/s)^{2}}{2} + 0] + 0[/tex]
q = 6597.711 kJ
Thus, we can conclude that heat transferred through the coil per unit mass of water is 6597.711 kJ.
