Chapter 20, Problem 4SP

(a)

To determine

The heat energy to be added to the water to increase the temperature.

Answer to Problem 4SP

The heat energy to be added to the water to increase the temperature is $8.37\times {10}^5\text{J}$.

Explanation of Solution

Given info: The mass of water is $2000\text{g}$, the initial temperature is $0\text{°C}$ and the final temperature is $100\text{°C}$.

Write the expression for the amount of heat required to raise the temperature of a substance.

$\begin{array}{c}Q=mc\Delta T\\ =mc\left(T_2-T_1\right)\end{array}$

Here,

$Q$ is the heat energy

$m$ is the mass of the substance

$c$ is the specific heat of the substance

$\Delta T$ is the increase in temperature

$T_1$ is the initial temperature

$T_2$ is the final temperature

Substitute $2000\text{g}$ for $m$, $1\text{cal/g°}\cdot \text{C}$ for $c$, $100\text{°C}$ for $T_2$ and $0\text{°C}$ for $T_1$ to find the heat required to raise the temperature of water, $Q$.

$\begin{array}{c}Q=\left(2000\text{g}\right)\left(1\text{cal/g°}\cdot \text{C}\right)\left(100\text{°C}-0\text{°C}\right)\\ =2\times {10}^5\text{cal}\\ =2\times {10}^5\text{cal}\times \frac{4.186\text{J}}{1\text{cal}}\\ =8.37\times {10}^5\text{J}\end{array}$

Conclusion:

Therefore, the heat energy to be added to the water to increase the temperature is $8.37\times {10}^5\text{J}$.

(b)

To determine

The increase in mass of water due to the addition of energy.

Answer to Problem 4SP

The increase in mass of water due to the addition of energy is $9.3\times {10}^{-12}\text{kg}$.

Explanation of Solution

Given info: The energy added to the water is $8.37\times {10}^5\text{J}$.

Write the expression for the mass-energy equivalence.

$E_0=m{c}^2$

Here,

$E_0$ is the energy

$m$ is the mass

$c$ is the speed of light

Solve for $m$.

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

Substitute $8.37\times {10}^5\text{J}$ for $E_0$ and $3\times {10}^8\text{m/s}$ for $c$ to find the mass equivalent to the added energy.

$\begin{array}{c}m=\frac{8.37\times {10}^5\text{J}}{{\left(3\times {10}^8\text{m/s}\right)}^2}\\ =9.3\times {10}^{-12}\text{kg}\end{array}$

Conclusion:

Therefore, the increase in mass of water due to the addition of energy is $9.3\times {10}^{-12}\text{kg}$.

(c)

To determine

To compare the mass increased upon energy addition to the original mass of water and whether it is measurable or not.

Answer to Problem 4SP

The increased mass of water is about $1$ part in $50$ billion and it is insignificant and not measurable.

Explanation of Solution

Given info: The original mass of water is $2\text{kg}$ and the increase in mass is $9.3\times {10}^{-12}\text{kg}$.

The ratio of the increased mass to the original mass of water is obtained as,

$\frac{9.3\times {10}^{-12}\text{kg}}{2\text{kg}}=\frac{1}{2\times {10}^{11}}$

Hence, the increased mass of water is about $1$ part in $50$ billion. This much small increase in mass is not measurable and it is insignificant.

Conclusion:

Therefore, the increased mass of water is about $1$ part in $50$ billion and it is insignificant and not measurable.

(d)

To determine

The kinetic energy possible to produce from the original mass of water.

Answer to Problem 4SP

The kinetic energy possible to produce from the original mass of water is $1.8\times {10}^{17}\text{J}$.

Explanation of Solution

Given info: The original mass of water is $2\text{kg}$.

Write the expression for the mass-energy equivalence.

$E_0=m{c}^2$

Substitute $2\text{kg}$ for $m$ and $3\times {10}^8\text{m/s}$ for $c$ to find the corresponding energy.

$\begin{array}{c}{E}_0=\left(2\text{kg}\right){\left(3\times {10}^8\text{m/s}\right)}^2\\ =1.8\times {10}^{17}\text{J}\end{array}$

Hence the kinetic energy possible to produce from the original mass of water is $1.8\times {10}^{17}\text{J}$

Conclusion:

Therefore, the kinetic energy possible to produce from the original mass of water is $1.8\times {10}^{17}\text{J}$.