Question

Take $C_{ice}=0.53cal/g{-}^{\circ }C$, $C_{water}=1.0cal/g{-}^{\circ }C$, $(L_f{)}_{water}=80cal/g$ and $(L_v{)}_{water}=529cal/g$. A nuclear power plant generates $500MW$ of waste heat that must be carried away by water pumped from a lake. If the water temperature is to rise ${10}^{\circ }C$, what is the required flow rate in $kg/s$?

• A. $1.2\times {10}^{4}kg/s$
• B. $1.2\times {10}^{2}kg/s$
• C. $1.2\times {10}^{5}kg/s$
• D. $1.2\times {10}^{3}kg/s$

Answer 1

The correct option is A $1.2\times {10}^{4}kg/s$

Using $Q=M_w{C}_w\Delta T$

$\therefore$ Power $P=\frac{Q}{t}=\frac{M_w}{t}{C}_w\Delta T$

$⟹$ $P=f{C}_w\Delta T$ where $f$ is the flow rate of water

Given : $C_w=1.0cal/g^{o}C=4200J/k{g}^{o}C$

$\Delta T={10}^{o}C$

$\therefore$ $500\times {10}^6=f\times 4200\times 10$

$⟹f=1.2\times {10}^4$ $kg/s$