Chapter 20, Problem 20.8OQ

Beryllium has roughly one-half the specific heat of water (H₂O). Rank the quantities of energy input required to produce the following changes from the largest to the smallest. In your ranking, note any cases of equality, (a) raising the temperature of 1 kg of H₂O from 20°C to 26ºC (b) raising the temperature of 2 kg of H₂O from 20ºC to 23°C (c) raising the temperature of 2 kg of H₂O from 1ºC to 4°C (d) raising the temperature of 2 kg of beryllium from — 1°C to 2°C (e) raising the temperature of 2 kg of H₂O from -1°C to 2°C

To determine

The rank from the largest to smallest according to the required energy input.

Answer to Problem 20.8OQ

The rank from the largest to smallest according to the required energy input is $Q_{\text{e}}>Q_{\text{a}}=Q_{\text{b}}=Q_{\text{c}}>Q_{\text{d}}$.

Explanation of Solution

Given info: The specific heat of the Beryllium is one half of the specific heat of the water.

Write the expression to relate the specific heat of the Beryllium to the specific heat of the water.

$c_{\text{b}}=\frac{c_{\text{w}}}{2}$

Here,

$c_{\text{b}}$ is the specific heat of the Beryllium.

$c_{\text{w}}$ is the specific heat of the water.

The specific heat of water is $4186\text{J}/\text{kg}\cdot °\text{C}$.

Substitute $4186\text{J}/\text{kg}\cdot °\text{C}$ for $c_{\text{w}}$ in the above equation.

$\begin{array}{c}{c}_{\text{b}}=\frac{4186\text{J}/\text{kg}\cdot °\text{C}}{2}\\ =2093\text{J}/\text{kg}\cdot °\text{C}\end{array}$

Thus, the specific heat of the Beryllium is $2093\text{J}/\text{kg}\cdot °\text{C}$.

The change in the temperature is calculated as,

$\Delta T=T_2-T_1$

Here,

$T_1$ is the initial temperature of the material.

$T_2$ is the final temperature of the material.

$\Delta T$ is the change in temperature.

The amount of energy required for the rise in temperature of water is,

$Q=cm\Delta T$

Here,

$m$ is the mass of the substance.

$Q$ is the amount of the energy required for the rise in temperature.

$c$ is the specific heat of the substance.

Substitute $\left(T_2-T_1\right)$ for $\Delta T$ in the above equation.

$Q=cm\left(T_2-T_1\right)$ (1)

For case (a),

The mass of $H_{2}O$ is $1.00\text{kg}$, the initial temperature is $20°\text{C}$ and the final temperature is $26°\text{C}$.

Substitute $1.00\text{kg}$ for $m$, $4186\text{J}/\text{kg}\cdot °\text{C}$ for $c$, $20°\text{C}$ for $T_1$ and $26°\text{C}$ for $T_2$ in the equation (1).

$\begin{array}{c}{Q}_{\text{a}}=4186\text{J}/\text{kg}\cdot °\text{C}\left(1.00\text{kg}\right)\left(26°\text{C}-20°\text{C}\right)\\ =25116\text{J}\\ \approx 2.51\times {10}^4\text{J}\end{array}$

Thus, the energy required in case (a) is $2.51\times {10}^4\text{J}$.

For case (b),

The mass of $H_{2}O$ is $2.00\text{kg}$, the initial temperature is $20°\text{C}$ and the final temperature is $23°\text{C}$.

Substitute $2.00\text{kg}$ for $m$, $4186\text{J}/\text{kg}\cdot °\text{C}$ for $c$, $20°\text{C}$ for $T_1$ and $23°\text{C}$ for $T_2$ in the equation (1).

$\begin{array}{c}{Q}_{\text{b}}=4186\text{J}/\text{kg}\cdot °\text{C}\left(2.00\text{kg}\right)\left(23°\text{C}-20°\text{C}\right)\\ =4186\text{J}/\text{kg}\cdot °\text{C}\left(2.00\text{kg}\right)\left(3°\text{C}\right)\\ =25116\text{J}\\ \approx 2.51\times {10}^4\text{J}\end{array}$

Thus, the energy required in case (b) is $2.51\times {10}^4\text{J}$.

For case (c),

The mass of $H_{2}O$ is $2.00\text{kg}$, the initial temperature is $1°\text{C}$ and the final temperature is $4°\text{C}$.

Substitute $2.00\text{kg}$ for $m$, $4186\text{J}/\text{kg}\cdot °\text{C}$ for $c$, $1°\text{C}$ for $T_1$ and $4°\text{C}$ for $T_2$ in the equation (1).

$\begin{array}{c}{Q}_{\text{c}}=4186\text{J}/\text{kg}\cdot °\text{C}\left(1.00\text{kg}\right)\left(4°\text{C}-1°\text{C}\right)\\ =4186\text{J}/\text{kg}\cdot °\text{C}\left(1.00\text{kg}\right)\left(3°\text{C}\right)\\ =25116\text{J}\\ \approx 2.51\times {10}^4\text{J}\end{array}$

Thus, the energy required in case (c) is $2.51\times {10}^4\text{J}$.

For case (d),

The mass of beryllium is $2.00\text{kg}$, the initial temperature is $-1°\text{C}$ and the final temperature is $2°\text{C}$, the specific heat of beryllium is $2093\text{J}/\text{kg}\cdot °\text{C}$.

Substitute $2.00\text{kg}$ for $m$, $2093\text{J}/\text{kg}\cdot °\text{C}$ for $c$, $-1°\text{C}$ for $T_1$ and $2°\text{C}$ for $T_2$ in the equation (1).

$\begin{array}{c}{Q}_{\text{d}}=2093\text{J}/\text{kg}\cdot °\text{C}\left(2.00\text{kg}\right)\left(2°\text{C}-\left(-1°\text{C}\right)\right)\\ =2093\text{J}/\text{kg}\cdot °\text{C}\left(2.00\text{kg}\right)\left(3°\text{C}\right)\\ =12558\text{J}\\ \approx 1.25\times {10}^4\text{J}\end{array}$

Thus, the energy required in case (d) is $1.25\times {10}^4\text{J}$.

For case (e),

The mass of $H_{2}O$ is $2.00\text{kg}$, the initial temperature is $-1°\text{C}$ and the final temperature is $2°\text{C}$, the specific heat of $H_{2}O$ is $4186\text{J}/\text{kg}\cdot °\text{C}$.

For the temperature in negative $H_{2}O$ is in the ice form, the specific heat of the ice is $2090\text{J}/\text{kg}\cdot °\text{C}$.

Substitute $2.00\text{kg}$ for $m$, $2090\text{J}/\text{kg}\cdot °\text{C}$ for $c$, $-1°\text{C}$ for $T_1$ and $0°\text{C}$ for $T_2$ in the equation (1).

$\begin{array}{c}{Q^{\prime }}_{\text{e}}=2090\text{J}/\text{kg}\cdot °\text{C}\left(2.00\text{kg}\right)\left(0°\text{C}-\left(-1°\text{C}\right)\right)\\ =2090\text{J}/\text{kg}\cdot °\text{C}\left(2.00\text{kg}\right)\left(1°\text{C}\right)\\ =4180\text{J}\\ =\text{0}\text{.0418}\times {\text{10}}^5\text{J}\end{array}$

From temperature $0°\text{C}$ to $0°\text{C}$ the ice melts and the energy required for melting the heat is,

${Q^{″}}_{\text{e}}=m\left(L_{\text{ice}}\right)$

Here,

$L_{\text{ice}}$ is the latent heat of fusion of ice.

${Q^{″}}_{\text{e}}$ is the energy required for melting.

The latent heat of fusion of ice is $3.33\times {10}^5\text{J}/\text{kg}$.

Substitute $2.00\text{kg}$ for $m$ and $3.33\times {10}^5\text{J}/\text{kg}$ for $L_{\text{ice}}$ in the above equation.

$\begin{array}{c}{Q^{″}}_{\text{e}}=2.00\text{kg}\left(3.33\times {10}^5\text{J}/\text{kg}\right)\\ =6.66\times {10}^5\text{J}\end{array}$

From $0°\text{C}$ to $2°\text{C}$ the $H_{2}O$ is in liquid form.

Substitute $2.00\text{kg}$ for $m$, $4186\text{J}/\text{kg}\cdot °\text{C}$ for $c$, $0°\text{C}$ for $T_1$ and $2°\text{C}$ for $T_2$ in the equation (1) to find the required energy in case of liquid form.

$\begin{array}{c}{Q^{‴}}_{\text{e}}=4186\text{J}/\text{kg}\cdot °\text{C}\left(2.00\text{kg}\right)\left(2°\text{C}-\left(0°\text{C}\right)\right)\\ =4186\text{J}/\text{kg}\cdot °\text{C}\left(2.00\text{kg}\right)\left(2°\text{C}\right)\\ =16744\text{J}\\ \approx 0.167\times {10}^4\text{J}\end{array}$

The total energy required in case (e) is,

$Q_{\text{e}}={Q^{\prime }}_{\text{e}}+{Q^{″}}_{\text{e}}+{Q^{‴}}_{\text{e}}$

Substitute $\text{0}\text{.0418}\times {\text{10}}^5\text{J}$ for ${Q^{\prime }}_{\text{e}}$, $6.66\times {10}^5\text{J}$ for ${Q^{″}}_{\text{e}}$ and $16744\text{J}$ for ${Q^{‴}}_{\text{e}}$ in the above equation.

$\begin{array}{c}{Q}_{\text{e}}=\text{0}\text{.0418}\times {\text{10}}^5\text{J}+6.66\times {10}^5\text{J}+16744\text{J}\\ =6.87\times {\text{10}}^5\text{J}\end{array}$

Thus, the energy required in case (e) is $6.87\times {\text{10}}^5\text{J}$.

Conclusion:

Therefore, the rank from the largest to smallest according to the required energy input is $Q_{\text{e}}>Q_{\text{a}}=Q_{\text{b}}=Q_{\text{c}}>Q_{\text{d}}$.