Chapter 12, Problem 49A

To determine

Mass of water used to cool the engine.

Answer to Problem 49A

Mass of water used $m_2=15.19\text{kg}$

Explanation of Solution

Given:

Mass of car, $m_1=2.5\times {10}^2\text{kg}$

Engine temperature when shut off, $T_2=35°\text{C}$

Air temperature, $T_1=10°\text{C}$

Heat energy, $Q=4.4\times {10}^6\text{J}$

Formula Used:

Specific heat formula,

  $Q=m_1{C}_1\Delta T_1+m_2{C}_2\Delta T_2$

Here, $Q$ is heat energy, $m_1$ is mass of car, $C_1$ is specific heat of air, and $\Delta T_1$ is change in temperature, $C_2$ is specific heat of water, and $\Delta T_2$ is change in temperature

Calculation:

From the above formula, mass of water $m_2$ is

  $m_2=\frac{Q-m_1{C}_1\Delta T_1}{C_2\Delta T_2}$

As specific heat of water is $C_2=4180\text{J/kg}\cdot \text{K}$ and specific heat of air is $C_1=450\text{J/kg}\cdot \text{K}$

Car engine and water are at same temperature..

Substitute the given values to find the mass,

  $\begin{array}{c}{m}_2=\frac{4.4\times {10}^6-\left(2.5\times {10}^2\right)\times 450\times \left(35-10\right)}{4180\times \left(35-10\right)}\\ =15.19\text{kg}\end{array}$

Conclusion:

Mass of water required to cool the engine is $m_2=15.19\text{kg}$ .