Question

A 44.0-g sample of an unknown metal at ${99.0}^{\circ }\mathrm{C}$ was placed in a constant-pressure calorimeter containing $80.0\mathrm{g}$ of water at ${24.0}^{\circ }\mathrm{C}$. The final temperature of the system was found to be ${28.4}^{\circ }\mathrm{C}$. Calculate the specific heat of the metal. (The heat capacity of the calorimeter is $12.4\mathrm{J}/{}^{\circ }\mathrm{C}.$

Short answer

The specific heat of the metal is approximately $0.491J/(g{}^{\circ }C)$.

Step-by-step solution

Understand the Heat Transfer

In a calorimetry problem like this one, the heat lost by the metal is equal to the heat gained by the water and the calorimeter. This follows from the principle of conservation of energy.

Write the Heat Transfer Equations

The heat lost by the metal is given by $q_{extmetal}=m_{extmetal}imes{c}_{extmetal}imes(T_{extfinal}-T_{extinitial,metal})$. The heat gained by the water is given by $q_{extwater}=m_{extwater}imes{c}_{extwater}imes(T_{extfinal}-T_{extinitial,water})$. The heat absorbed by the calorimeter is $q_{extcalorimeter}=C_{extcalorimeter}imes(T_{extfinal}-T_{extinitial,calorimeter})$. Here, $c_{extwater}=4.18\mathrm{J}/(\mathrm{g}{}^{\mathrm{ext}\circ }\mathrm{C})$ and $C_{extcalorimeter}$ is provided.

Set up the Equation for Conservation of Energy

According to the conservation of energy: $-q_{extmetal}=q_{extwater}+q_{extcalorimeter}$ This can be rearranged to: $$m_{extmetal}imes{c}_{extmetal}imes(T_{extfinal}-T_{extinitial,metal})=-(m_{extwater}imes{c}_{extwater}\times (T_{extfinal}-T_{extinitial,water})+C_{extcalorimeter}imes(T_{extfinal}-T_{extinitial,calorimeter}))$$

Plug in the Known Values

Substitute the given values:- $m_{extmetal}=44.0g$- $T_{extinitial,metal}={99.0}^{ext\circ }C$, $T_{extfinal}={28.4}^{ext\circ }C$- $m_{extwater}=80.0g$- $T_{extinitial,water}={24.0}^{ext\circ }C$- $C_{extcalorimeter}=12.4J{/}^{ext\circ }C$- $c_{extwater}=4.18J/(g{\cdot }^{ext\circ }C)$.

Calculate the Heat Gained by the Water and Calorimeter

Calculate each part:$$q_{extwater}=80.0g\times 4.18J/(g{\cdot }^{ext\circ }C)\times (28.4-24.0{)}^{\circ }C=80.0\times 4.18\times 4.4=1470.08J$$$$q_{extcalorimeter}=12.4J{/}^{\circ }C\times (28.4-24.0{)}^{\circ }C=54.56J$$

Calculate Total Heat Absorbed and Metal's Heat Loss

Sum the heat gained by the water and the calorimeter:$$q_{extwater}+q_{extcalorimeter}=1470.08J+54.56J=1524.64J$$Now, equal this to the negative of the heat lost by the metal: $$q_{extmetal}=-1524.64J$$

Solve for Specific Heat of the Metal

Use the rearranged conservation of energy equation and solve for $c_{extmetal}$:$$44.0g\times c_{extmetal}\times (28.4-99.0{)}^{\circ }C=-1524.64J$$Simplifying:$$44.0\times c_{extmetal}\times (-70.6)=-1524.64$$Divide by the calculated product to find $c_{extmetal}$:$$c_{extmetal}=\frac{1524.64}{3106.4}\approx 0.491J/(g{\cdot }^{\circ }C)$$

Verify and Reflect

Reassure that the calculated specific heat is reasonable by comparing it with known specific heat values for metals, confirming the methodology and calculations are correct. This helps ensure the solution is validated.

Key concepts

Specific Heat Capacity

Specific heat capacity is a crucial concept in calorimetry. It refers to the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius.
Different materials have varying specific heat capacities.
For example, water has a specific heat capacity of 4.18 J/(g°C), which is quite high. This means it takes a lot of heat for water to change temperature.
This property is essential in calorimetry when determining the specific heat of an unknown substance, like a metal in our exercise.

The general formula used is:

• $$q=mimescimes\mathrm{\Delta }T$$
• Where:
• $q$ is the heat energy (in Joules).
• $m$ is the mass (in grams).
• $c$ is the specific heat capacity (in J/(g°C)).
• $\mathrm{\Delta }T$ is the change in temperature.

This equation helps to calculate the specific heat of an unknown metal by substituting in the other known values.

Conservation of Energy

The principle of conservation of energy states that energy cannot be created or destroyed; it can only be transferred or transformed.
This concept is pivotal in calorimetry problems where we assume the system is isolated from external influences.
It means that the heat energy lost by a hot object must equal the heat energy gained by a cooler object.

In calorimetry, the conservation of energy can be mathematically expressed as:

• $q_{\text{metal}}+q_{\text{water}}+q_{\text{calorimeter}}=0$

This signifies:

• The energy lost by the unknown metal (negative) is equal to the energy gained by the water and the calorimeter.

This fundamental principle allows us to solve for unknown quantities in calorimetric calculations, such as the specific heat of the metal in the exercise.

Heat Transfer

Heat transfer is the movement of thermal energy from one object or substance to another.
It is driven by the temperature difference between the two, flowing from the hotter to the cooler body.

In the context of the calorimetry exercise, heat transfer occurs when the hot metal is placed in cooler water and the calorimeter.
Heat energy from the metal is transferred to the water and the calorimeter until thermal equilibrium is achieved - the point where their temperatures equalize.

The efficiency of this transfer often depends on:

• The temperature difference (greater differences lead to faster energy transfer).
• The thermal properties of the substances involved, such as thermal conductivity.
• The surface area in contact and the duration of the heat transfer process.

Understanding heat transfer enables the calculation of the specific heat capacity of unknown substances by analyzing how heat flows in the system.

Unknown Metal Identification

Identifying an unknown metal involves determining its specific heat capacity and comparing it with known values for various metals.
Once we calculate the specific heat using calorimetry, it can serve as a fingerprint to help identify the material.

To identify the metal in the exercise, you:

• First calculate the specific heat capacity from the calorimetry experiment.
• Next, compare the calculated value to a database or table of specific heat capacities of known metals.

For instance, if the calculated specific heat matches that of aluminum, you can infer that the unknown metal is likely aluminum.

This process is incredibly useful in various applications, from material science research to quality control in manufacturing.