Question

A nuclear reactor core must stay at or below 95 °C to remain in good working condition. Cool water at a temperature of 10 °C is used to cool the reactor. If the reactor emits 23746 kJ of energy each hour, how many grams of water must circulate each hour in order to keep the reactor at or below 95 °C?

Answer 1

`\large \boxed{\text{67 000 g}}`

Step-by-step explanation:

This is a problem in calorimetry — the measurement of the quantities of heat that flow from one object to another.

It is based on the Law of Conservation of Energy — Energy can be transformed from one type to another, but it cannot be destroyed or created.

If heat flows out of the reactor (negative), the same amount of heat must flow into the water (positive).

Since there is no change in total energy,

heat₁ + heat₂ = 0

The symbol for the quantity of heat transferred is q, so we can rewrite the word equation as

q₁ + q₂ = 0

The formula for the heat absorbed or released by an object is

q = mCΔT, where

m = the mass of the sample

C = the specific heat capacity of the sample, and

ΔT = T_f - T_i = the change in temperature

1. Equation

There are two heat flows in this problem,

heat released by reactor + heat absorbed by water = 0

q₁ + q₂ = 0

q₁ + m₂C₂ΔT₂ = 0

2. Data:

q₁ = -23 746 kJ

m₂ = ?; C₂ = 4.184 J°C⁻¹g⁻¹; T_f = 95 °C; T_i = 10 °C

3. Calculations

(a) Convert kilojoules to joules

`q_(1) = -\text{23 746 kJ} * \frac{\text{1000 J}}{\text{1 kJ}} = -\text{23 746 000 J}`

(b) ΔT

ΔT₂ = T_f - T_i = 95 °C - 10 °C = 85 °C

(c) m₂

`\begin{array}{rcl}q_(1) + q_(2) & = & 0\\\text{-23 746 000 J} + m_(2) * 4.184 \text{ J$^(\circ)$C$^(-1)$g$^(-1)$} * 85 \, ^(\circ)\text{C} & = & 0\\\text{-23 746 000 J} + 356m_(2) \text{J$\cdot$g}^(-1) & = & 0\\356m_(2) \text{g}^(-1) & = & 23746000\\m_2&=& \frac{23746000}{\text{356 g}^(-1)}\\\\ & = & \textbf{67000 g}\\\end{array}\\`

`\text{You must circulate $\large \boxed{\textbf{67 000 g}}$ of water each hour.}`