Chapter 20, Problem 4SP

(a)

To determine

The heat energy added to raise the temperature of water from $0°\text{C}$ to $100°\text{C}$.

Answer to Problem 4SP

The heat energy added to raise the temperature of water from $0°\text{C}$ to $100°\text{C}$ is $1.05\times {10}^6\text{ J}$.

Explanation of Solution

Given info: The mass of the water is $2.5\text{ kg}$. The specific heat of water is $4186\text{J}/\text{kg}\cdot \text{K}$. The change in temperature is $100°\text{C}=\text{100 K}$.

Write the expression to find the heat energy added to raise the temperature of water from $0°\text{C}$ to $100°\text{C}$.

$Q=mc\Delta t$

Here,

$Q$ is the heat energy

$m$ is the mass of the water

$c$ is the specific heat of water

$\Delta t$ is the change in temperature

Substitute $100\text{ K}$ for $\Delta t$, $4186\text{J}/\text{kg}\cdot \text{K}$ for $c$ and $2.5\text{ kg}$ for $m$ in the above equation.

$\begin{array}{c}Q=\left(2.5\text{ kg}\right)\left(4186\text{J}/\text{kg}\cdot \text{K}\right)\left(100\text{ K}\right)\\ =1.05\times {10}^6\text{ J}\end{array}$

Conclusion:

Therefore, the heat energy added to raise the temperature of water from $0°\text{C}$ to $100°\text{C}$ is $1.05\times {10}^6\text{ J}$.

(b)

To determine

The amount of water increasing in the process.

Answer to Problem 4SP

The amount of water increasing in the process is $1.16\times {10}^{-11}\text{ kg}$.

Explanation of Solution

Write the expression to find mass of the water increasing in the process.

$m=\frac{E_0}{c^2}$

Here,

$E_0$ is the energy

$c$ is the speed of light

$m$ is the amount of water

Substitute $1.05\times {10}^6\text{ J}$ for $E_0$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in the above equation.

$\begin{array}{c}m=\frac{1.05\times {10}^6\text{ J}}{{\left(3\times {10}^8\text{m}/\text{s}\right)}^2}\\ =1.16\times {10}^{-11}\text{ kg}\end{array}$

Conclusion:

Therefore, the amount of water increasing in the process is $1.16\times {10}^{-11}\text{ kg}$.

(c)

To determine

The comparison of the increase in mass with the original mass.

Answer to Problem 4SP

The increase in mass is negligible when compared with the original mass.

Explanation of Solution

When we consider the expression of the ratio of the original mass to the increase in mass, the ratio becomes $2.5\text{ kg : 1}\text{.16}\times {\text{10}}^{-11}\text{ kg}$. This is very negligible.

This can be reduced to $1$ part in $200\text{ billion}$ which is very negligible. This shows the insignificance of the increase in mass in this case.

Conclusion:

Therefore, the increase in mass is negligible when compared with the original mass.

(d)

To determine

The conversion of original mass into kinetic energy.

Answer to Problem 4SP

The conversion of original mass into kinetic energy gives $2.25\times {10}^{17}\text{ J}$.

Explanation of Solution

Write the expression for conversion of original mass into kinetic energy.

$KE=m{c}^2$

Here,

$m$ is the original mass

$c$ is the speed of light

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $2.5\text{ kg}$ for $m$ in the above equation.

$\begin{array}{c}KE=\left(2.5\text{ kg}\right){\left(3\times {10}^8\text{m}/\text{s}\right)}^2\\ =2.25\times {10}^{17}\text{ J}\end{array}$

Conclusion:

Therefore, the conversion of original mass into kinetic energy gives $2.25\times {10}^{17}\text{ J}$.