Problem 13 - Chemistry and Chemical Reactivity

Question

A \(182-\mathrm{g}\) sample of gold at some temperature was added to \(22.1 \mathrm{g}\) of water. The initial water temperature was \(25.0^{\circ} \mathrm{C},\) and the final temperature was \(27.5^{\circ} \mathrm{C} .\) If the specific heat capacity of gold is \(0.128 \mathrm{J} / \mathrm{g} \cdot \mathrm{K},\) what was the initial temperature of the gold sample?

Step-by-Step Solution

Answer

The initial temperature of the gold sample was approximately 258.06°C.

Step 1: Understand the Concept of Heat Exchange

This problem involves the concept of heat transfer, where the heat lost by the gold will equal the heat gained by the water. The relationship can be stated as: \[ q_{ ext{gold}} = -q_{ ext{water}} \]Here, \( q \) represents heat in joules. Since no energy is lost to the surroundings, the sum of heat exchange is zero.

Step 2: Write the Formula for Heat Transfer

The heat transfer for a substance can be calculated using: \[ q = mc\Delta T \]where \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. For water, \( c = 4.18 \) J/g·K.

Step 3: Calculate Heat Gained by Water

Use the formula from Step 2 for water:\[ q_{ ext{water}} = m_{ ext{water}} \, c_{ ext{water}} \, \Delta T_{ ext{water}} \]\[ q_{ ext{water}} = (22.1\, \text{g}) (4.18 \, \text{J/g} \cdot \text{K}) (27.5 - 25.0) \, \text{K} \]Calculate to get \( q_{ ext{water}} = 230.13 \) J.

Step 4: Represent Heat Lost by Gold

Since the heat lost by gold is equal to the heat gained by water (with opposite sign):\[ q_{ ext{gold}} = -q_{ ext{water}} = -230.13 \, \text{J} \]

Step 5: Solve for Initial Temperature of Gold

Using the heat transfer formula for gold, \[ q_{ ext{gold}} = m_{ ext{gold}} \, c_{ ext{gold}} \, \Delta T_{ ext{gold}} \]Substitute known values:\[ -230.13 = (182 \, \text{g}) (0.128 \, \text{J/g} \cdot \text{K}) (27.5 - T_i) \]Solving for \( T_i \), we find:\[ T_i =\frac{ -230.13 }{ - (182 \, \text{g}) (0.128 \, \text{J/g} \cdot \text{K}) } + 27.5 \]\[ T_i = 230.56 + 27.5 = 258.06^{ullet} \text{C} \]

Step 6: Final Calculation and Result

Perform the above calculation for \( T_i \):\[ T_i \approx 258.06^{ullet} \text{C} \]This is the initial temperature of the gold sample.

Key Concepts

Specific Heat Capacity

Specific heat capacity is a crucial concept in understanding heat transfer processes. It denotes the amount of heat required to change a substance's temperature by one degree Celsius or Kelvin, per unit mass. For example, specific heat capacity helps in determining how different materials store and transfer heat energy.
- **Formula**: The specific heat capacity is denoted by the symbol \( c \) and is used in the equation \( q = mc\Delta T \), where \( q \) represents the heat energy transferred, \( m \) is the mass, and \( \Delta T \) is the change in temperature.
- **Typical Values**: Materials have different specific heat capacities. Water, for example, has a high specific heat capacity of 4.18 J/g·K, which means it can absorb a lot of heat before its temperature changes significantly. Gold, on the other hand, has a lower specific heat capacity of 0.128 J/g·K, making it heat up or cool down quicker with the same amount of heat exchange.
The concept is important when calculating how much heat an object can gain or lose, and it can determine the effectiveness of materials in thermal management applications.

Temperature Change

Understanding temperature change is fundamental in solving calorimetry problems. It refers to the difference in temperature that a material undergoes during a heating or cooling process. This is not only essential in calculations but also in understanding how energy is transferred between substances.
- **Measuring Change**: The change in temperature, \( \Delta T \), is calculated by subtracting the initial temperature from the final temperature: \( \Delta T = T_f - T_i \). This tells us how much the temperature has increased or decreased during the process.
- **Practical Example**: In the given exercise, the water’s temperature changed from 25.0°C to 27.5°C, giving a \( \Delta T \) of 2.5°C. This piece of information is crucial when calculating the heat gained or lost.
Temperature change is an observable outcome of heat transfer, and is a key factor in determining the effectiveness of thermal exchange in physical processes.

Calorimetry

Calorimetry is the science of measuring the amount of heat involved in a chemical or physical process. This process is widely used in many scientific studies and practical applications, such as determining the heat capacity of materials, understanding metabolic rates, and more.
- **Fundamental Principle**: Calorimetry is based on the principle that energy in a closed system is conserved. This means the heat lost by one substance must be equal to the heat gained by another, except for any minor losses to the surroundings, which are often negligible for classroom calculations.
- **Equation to Use**: The main equation of calorimetry is \( q_{ ext{lost}} = -q_{ ext{gained}} \). This implies that if one body loses heat, another must gain an equivalent amount of energy. In the problem with gold and water, this equation is key to finding unknown temperatures or masses.
- **Real-Life Application**: A calorimeter, a tool used in calorimetry, provides an insulated environment to measure the heat transfer between substances accurately, allowing for precise calculations of heat flow and temperature change.
Calorimetry not only aids in understanding heat exchange but also mimics real-world thermal interactions, providing a comprehensive view of energy transfers.