Chapter 20, Problem 13P

(a)

To determine

The specific heat of unknown sample.

Answer to Problem 13P

The specific heat of unknown sample is $1822\text{J}/\text{kg}\cdot \text{°C}$.

Explanation of Solution

Given info: The mass of the calorimeter is $100\text{g}$, the mass of water in calorimeter is $250\text{g}$, the equilibrium temperature of water and calorimeter is $10.0\text{°C}$, the mass of copper block is $50.0\text{g}$, the temperature of copper block is $80.0°\text{C}$, the mass of other block is $70.0\text{g}$ and the temperature of other block is $100°\text{C}$. The final temperature of the system is $20.0°\text{C}$.

Formula to calculate the heat lost by cooper block is,

$Q_{\text{Cu}}=m_{\text{Cu}}{c}_{\text{Cu}}\left(T_{\text{Cu}}-T\right)$

Here,

$Q_{\text{Cu}}$ is the heat lost by copper block.

$m_{\text{Cu}}$ is the mass of the copper block.

$c_{\text{Cu}}$ is the specific heat of copper.

$T_{\text{Cu}}$ is the temperature of copper block.

$T$ is the final temperature of the system.

Formula to calculate the heat lost by unknown sample is,

$Q_{\text{s}}=m_{\text{s}}{c}_{\text{s}}\left(T_{\text{s}}-T\right)$

Here,

$Q_{\text{s}}$ is the heat lost by sample.

$m_{\text{s}}$ is the mass of the sample.

$c_{\text{s}}$ is the specific heat of sample.

$T_{\text{s}}$ is the temperature of sample.

Formula to calculate the heat gained by the water is,

$Q_{\text{w}}=m_{\text{w}}{c}_{\text{w}}\left(T-T_{\text{w}}\right)$

Here,

$Q_{\text{w}}$ is the heat gained by water.

$m_{\text{w}}$ is the mass of water.

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

$T_{\text{w}}$ is the temperature of water.

Formula to calculate the heat gained by calorimeter is,

$Q_{\text{cal}}=m_{\text{cal}}{c}_{\text{Al}}\left(T-T_{\text{cal}}\right)$

Here,

$Q_{\text{cal}}$ is the heat gained by calorimeter.

$m_{\text{cal}}$ is the mass of calorimeter.

$c_{\text{Al}}$ is the specific heat of aluminum.

$T_{\text{cal}}$ is the temperature of calorimeter.

From the conservation of energy, heat lost is equal to heat gained.

$\begin{array}{c}{Q}_{\text{Cu}}+Q_{\text{s}}=Q_{\text{w}}+Q_{\text{cal}}\\ m_{\text{Cu}}{c}_{\text{Cu}}\left(T_{\text{Cu}}-T\right)+m_{\text{s}}{c}_{\text{s}}\left(T_{\text{s}}-T\right)=m_{\text{w}}{c}_{\text{w}}\left(T-T_{\text{w}}\right)+m_{\text{cal}}{c}_{\text{Al}}\left(T-T_{\text{cal}}\right)\\ m_{\text{s}}{c}_{\text{s}}\left(T_{\text{s}}-T\right)=m_{\text{w}}{c}_{\text{w}}\left(T-T_{\text{w}}\right)+m_{\text{cal}}{c}_{\text{Al}}\left(T-T_{\text{cal}}\right)-m_{\text{Cu}}{c}_{\text{Cu}}\left(T_{\text{Cu}}-T\right)\\ c_{\text{s}}=\frac{m_{\text{w}}{c}_{\text{w}}\left(T-T_{\text{w}}\right)+m_{\text{cal}}{c}_{\text{Al}}\left(T-T_{\text{cal}}\right)-m_{\text{Cu}}{c}_{\text{Cu}}\left(T_{\text{Cu}}-T\right)}{m_{\text{s}}\left(T_{\text{s}}-T\right)}\end{array}$

Substitute $250\text{g}$ for $m_{\text{w}}$, $100\text{g}$ for $m_{\text{cal}}$, $10.0°\text{C}$ for $T_{\text{w}}\text{and}{T}_{\text{cal}}$, $4186\text{J}/\text{kg}\cdot \text{°C}$ for $c_{\text{w}}$, $900\text{J}/\text{kg}\cdot \text{°C}$ for $c_{\text{Al}}$, $50.0\text{g}$ for $m_{\text{Cu}}$, $80.0°\text{C}$ for $T_{\text{Cu}}$, $387\text{J}/\text{kg}\cdot \text{°C}$ for $c_{\text{Cu}}$, $70.0\text{g}$ for $m_{\text{s}}$, $100°\text{C}$ for $T_{\text{s}}$ and $20.0°\text{C}$ for $T$ to find $c_{\text{s}}$.

$\begin{array}{c}{c}_{\text{s}}=\frac{\left[\begin{array}{l}250\text{g}\times 4186\text{J}/\text{kg}\cdot \text{°C}\times \left(20.0°\text{C}-10.0°\text{C}\right)+100\text{g}\times 900\text{J}/\text{kg}\cdot \text{°C}\left(20.0°\text{C}-10.0°\text{C}\right)\\ -50.0\text{g}\times 387\text{J}/\text{kg}\cdot \text{°C}\left(80.0°\text{C}-20.0°\text{C}\right)\end{array}\right]}{70.0\text{g}\left(100°\text{C}-20.0°\text{C}\right)}\\ =\frac{10465000\text{g}\cdot \text{°C}\times \text{J}/\text{kg}\cdot \text{°C}+900000\text{g}\cdot \text{°C}\times \text{J}/\text{kg}\cdot \text{°C}-1161000\text{g}\cdot \text{°C}\times \text{J}/\text{kg}\cdot \text{°C}}{70.0\text{g}\times 80.0°\text{C}}\\ =1822\text{J}/\text{kg}\cdot \text{°C}\end{array}$

Conclusion:

Therefore, the specific heat of unknown sample is $1822\text{J}/\text{kg}\cdot \text{°C}$.

(b)

To determine

The specific heat of unknown sample.

Answer to Problem 13P

The unknown sample may be beryllium.

Explanation of Solution

Given info: The mass of the calorimeter is $100\text{g}$, the mass of water in calorimeter is $250\text{g}$, the equilibrium temperature of water and calorimeter is $10.0\text{°C}$, the mass of copper block is $50.0\text{g}$, the temperature of copper block is $80.0°\text{C}$, the mass of other block is $70.0\text{g}$ and the temperature of other block is $100°\text{C}$. The final temperature of the system is $20.0°\text{C}$.

The specific heat of unknown sample is $1822\text{J}/\text{kg}\cdot \text{°C}$.

This value of specific heat is not given in the given table. So, it is difficult to make a definite identification of the sample. The closest value of this which is given in table is $1830\text{J}/\text{kg}\cdot \text{°C}$. Thus, the unknown sample may be beryllium.

Conclusion:

Therefore, the unknown sample may be beryllium.

(c)

To determine

To explain: The answer obtained in part (b).

Answer to Problem 13P

The specific heat may be defined as either the material having this amount of specific heat is not listed in the table or it may be an unknown alloy.

Explanation of Solution

The specific heat may be defined as the heat capacity per unit mass. It is the amount of heat per unit mass required to raise the temperature by $1°\text{C}$.

The value of specific heat obtained in part (b) does not correspond to any material given in table 19.1. So, either the material is not listed in the table or it may be an unknown alloy.

Conclusion:

Therefore, either the material having this amount of specific heat is not listed in the table or it may be an unknown alloy.