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A typical nuclear reactor generates 1,000 MW (1,000 MJ/s) of electrical energy. In doing so, it produces 2,000 MW of "waste heat" that must be removed from the reactor to keep it from melting down. Many reactors are sited next to large bodies of water so that they can use the water for cooling. Consider a reactor where the intake water is at 18° C. State regulations limit the temperature of the output water to 30° C so as to not harm aquatic organisms. How many liters of cooling water have to be pumped through the reactor each minute?

Answer 1

Answer:

There have to be pumped 100,646 liters of cooling water through the reactor each minute.

Explanation:

In a minute, the reactor produces 2,000MW*60s=120,000MJ of heat. We use the formula

$Q=mC\Delta T$

Where $Q$ is the transferred heat, $m$ is the mass of the water used, $C$ is the specific heat of water (C=4.1813 J/g°K) and $\Delta T$ is the change in temperature of the water.

If we solve for $m$ we get:

$m=\frac{Q}{C\Delta T}$

But we need the volume $V$ of the water, not its mass. So we have to use the concept of density. Density $\rho$ is equal to

$\rho=\frac{m}{V}$

So,

$m=\rho V$

We put this in the other equation, and get:

$V=\frac{Q}{C\Delta T \rho}$

Before we plug in the known values in the formula, we have to convert them to the correct units:

• $Q:$ 120,000MJ = 1.2*10¹¹J
• $C$ = 4.1813J/g°K
• $\Delta T:$ 30°C-18°C = 12°C = 285.15°K
• $\rho$ = 1,000g/L

Finally, we calculate the volume of water:

$V=\frac{1,2*10^{11} J}{(4.1813J/gK)(285.15K)(1,000g/L)} =100,646L$

So, there have to be pumped 100,646 liters of cooling water through the reactor each minute.