Answer:
[tex]\Delta t= 2962.395\,s\,(49.373\,min)[/tex]
Explanation:
Let assume that cooling water works under a pressure of 1 atmosphere. The time required to boil half of the water is determined by the First Law of Thermodynamics:
[tex]\dot Q \cdot \Delta t = m \cdot [c_{p,w}\cdot (T_{2}-T_{1})+h_{fg}][/tex]
[tex]\Delta t = \frac{m\cdot [c_{p,w}\cdot (T_{2}-T_{1})+h_{fg}]}{\dot Q}[/tex]
[tex]\Delta t = \frac{(2.25\times 10^{5}\,kg)\left[\left(4.186\,\frac{kJ}{kg\cdot ^{\textdegree}C} \right)\cdot (100\,^{\textdegree}C - 10\,^{\textdegree}C)+2256.5\,\frac{kJ}{kg} \right]}{200000\,kW}[/tex]
[tex]\Delta t= 2962.395\,s\,(49.373\,min)[/tex]
