Chapter 11, Problem 21P

(a)

To determine

The specific heat of unknown sample.

Answer to Problem 21P

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

Explanation of Solution

Given Info: Initial temperature of water and calorimeter is $10.0°\text{C}$ , mass of aluminum calorimeter is 0.100 kg, mass of water is 0.250 kg, mass of copper 50.0 g, mass of unknown block is 70.0 g, initial temperature of copper is $80.0°\text{C}$, initial temperature of the unknown block is $100°\text{C}$ and final temperature is $20.0°\text{C}$.

The heat lost by the copper and unknown block is equal to the heat gained by calorimeter and water.

Formula to calculate Heat gained by water is,

$Q_W=m_W{c}_W\left(T_f-T_{iW}\right)$

• $Q_W$ is the heat gained by water,
• $m_W$ is the mass of water,
• $c_W$ is the specific heat of water,
• $T_{iW}$ is the initial temperature of water,
• $T_f$ is the final temperature of water,

Formula to calculate Heat gained by aluminum calorimeter is,

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

• $Q_{Al}$ is the heat gained by aluminum calorimeter,
• $m_{\text{Al}}$ is the mass of calorimeter,
• $c_{\text{Al}}$ is the specific heat of aluminum,
• $T_{i\text{Al}}$ is the initial temperature of aluminum calorimeter,

Formula to calculate Heat lost by copper block is,

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

• $Q_{\text{Cu}}$ is the heat lost by copper block,
• $m_{\text{Cu}}$ is the mass of copper block,
• $c_{\text{Cu}}$ is the specific heat of copper block,
• $T_{i_{\text{Cu}}}$ is the initial temperature of copper block,

Formula to calculate Heat lost by unknown block is,

$Q_X=m_X{c}_X\left(T_f-T_{iX}\right)$

• $Q_X$ is the heat lost by unknown block,
• $m_X$ is the mass of unknown block,
• $c_X$ is the specific heat of the unknown block,
• $T_{iX}$ is the initial temperature of the unknown block,

The heat lost by the copper and unknown block is equal to the heat gained by calorimeter and water.

$Q_{\text{W}}+Q_{\text{Al}}=-\left(Q_{\text{Cu}}+Q_X\right)$

Use $m_{\text{Al}}{c}_{\text{Al}}\left(T_f-T_{i\text{Al}}\right)$ for $Q_{\text{Al}}$, $m_{\text{Cu}}{c}_{\text{Cu}}\left(T_f-T_{i_{\text{Cu}}}\right)$ for $Q_{\text{Cu}}$, $m_W{c}_W\left(T_f-T_{iW}\right)$ for $Q_W$, and $m_X{c}_X\left(T_f-T_{iX}\right)$ for $Q_X$ to rewrite in terms of $c_X$.

$\begin{array}{c}{m}_{\text{Al}}{c}_{\text{Al}}\left(T_f-T_{i\text{Al}}\right)+m_W{c}_W\left(T_f-T_{iW}\right)=-\left[m_{\text{Cu}}{c}_{\text{Cu}}\left(T_f-T_{i_{\text{Cu}}}\right)+m_X{c}_X\left(T_f-T_{iX}\right)\right]\\ -m_X{c}_X\left(T_f-T_{iX}\right)=m_{\text{Al}}{c}_{\text{Al}}\left(T_f-T_{i\text{Al}}\right)+m_W{c}_W\left(T_f-T_{iW}\right)+m_{\text{Cu}}{c}_{\text{Cu}}\left(T_f-T_{i_{\text{Cu}}}\right)\end{array}$

Substitute 0.250 kg for $m_W$, 0.100 kg for $m_{Al}$, $900\text{J}/\text{kg}\cdot \text{°C}$ for $c_{Al}$, $20.0°\text{C}$ for $T_f$ , $10.0°\text{C}$ for $T_{iW}$, $10.0°\text{C}$ for $T_{iAl}$, $4186\text{J}/\text{kg}\cdot \text{°C}$ for $c_W$, and $10.0°\text{C}$ for $T_{iW}$, 50.0 g for $m_{\text{Cu}}$, $387\text{J}/\text{kg}\cdot \text{°C}$ for $c_{\text{Cu}}$ , $80.0°\text{C}$ for $T_{i\text{Cu}}$, $100.0°\text{C}$ for $T_{iX}$, 70.0 g for $m_X$ to find $c_X$.

$\begin{array}{c}-\left[\left(70.0\text{g}\right)\left(\frac{1\text{kg}}{{10}^3\text{g}}\right)c_X\left(20.0°\text{C}-100.0°\text{C}\right)\right]=\left\{\begin{array}{l}\left(0.100\text{kg}\right)\left(900\text{J}/\text{kg}\cdot \text{°C}\right)\left(20.0°\text{C}-10.0°\text{C}\right)\\ +\left(0.250\text{kg}\right)\left(4186\text{J}/\text{kg}\cdot \text{°C}\right)\left(20.0°\text{C}-10.0°\text{C}\right)\\ +\left(50.0\text{g}\right)\left(\frac{1\text{kg}}{{10}^3\text{g}}\right)\left(387\text{J}/\text{kg}\cdot \text{°C}\right)\left(20.0°\text{C}-80.0°\text{C}\right)\end{array}\right\}\\ \left(5.6\text{kg}\cdot \text{°C}\right)c_X=10204\text{J}\\ c_X=1822\text{J}/\text{kg}\cdot \text{°C}\end{array}$

Conclusion:

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

(b)

To determine

The unknown material.

Explanation of Solution

The value of specific heat help to determine the substance.

Substance has their own value of specific heat. It is amount of heat required to raise the temperature of 1 kg of substance to $1°\text{C}$. The calculate value of specific heat resembles the specific heat of Beryllium.

Conclusion: Therefore, the unknown substance May be Beryllium

(c)

To determine

Explaining for part (b).

Explanation of Solution

If a person has more specific heat it requires more heat energy to raise its temperature. Specific heat helps in determining the substance.

Water has specific heat capacity of $4186\text{J}/\text{kg}\cdot \text{°C}$, when common salt is added to the water, the specific heat of the Answer changes. So, the value for specific heat obtained in part (a) can be specific heat of a metal or alloy. Hence, the substance is not exactly identified with the specific heat.

Conclusion: The substance could be Beryllium, alloy or any other material.