## Essential University Physics: Volume 1 (3rd Edition)

As $P_{out}=P_{in}-P$ We plug in the known values to obtain: $P_{out}=750-(750*.35)= 488 MW$ We know that $\Delta T=\frac{P_{out}}{mc}$ $\implies \Delta T=\frac{488 \times 10^6}{1.1\times 10^5\times 4184}=1.1 K$ As the temperature of river increases by 1.1 K, so the standard can be achieved by using river water for cooling.