Problem 46 - College Physics

Question

\(\bullet\) A laboratory technician drops an 85.0 g solid sample of unknown material at a temperature of \(100.0^{\circ} \mathrm{C}\) into a calorimeter. The calorimeter can is made of 0.150 \(\mathrm{kg}\) of copper and contains 0.200 \(\mathrm{kg}\) of water, and both the can and water are initially at \(19.0^{\circ} \mathrm{C}\) . The final temperature of the system is measured to be \(26.1^{\circ} \mathrm{C}\) . Compute the specific heat capacity of the sample. (Assume no heat loss to the surroundings.)

Step-by-Step Solution

Answer

The specific heat capacity of the sample is approximately 1.014 J/g°C.

Step 1: Set Up Energy Balance Equation

Start by writing the energy balance equation for the calorimeter. Since no heat is lost to the surroundings, the heat lost by the sample equals the heat gained by water and copper. We use the formula: \( q_{sample} + q_{water} + q_{copper} = 0 \). Here, \( q \) is the heat transferred, which is calculated as \( q = m \cdot c \cdot \Delta T \), where \( m \) is mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.

Step 2: Calculate Heat Gained by Water

Calculate the heat gained by the water using the formula \( q_{water} = m_{water} \cdot c_{water} \cdot (T_{final} - T_{initial}) \), where \( c_{water} = 4.18 \text{ J/g°C} \) is the specific heat capacity of water.\[ m_{water} = 0.200 \times 1000 = 200 \text{ g} \]\[ q_{water} = 200 \times 4.18 \times (26.1 - 19.0) \]\[ q_{water} = 200 \times 4.18 \times 7.1 = 5942 \text{ J} \]

Step 3: Calculate Heat Gained by Copper

Calculate the heat gained by the copper using the formula \( q_{copper} = m_{copper} \cdot c_{copper} \cdot (T_{final} - T_{initial}) \), where \( c_{copper} = 0.385 \text{ J/g°C} \).\[ m_{copper} = 0.150 \times 1000 = 150 \text{ g} \]\[ q_{copper} = 150 \times 0.385 \times (26.1 - 19.0) \]\[ q_{copper} = 150 \times 0.385 \times 7.1 = 410.175 \text{ J} \]

Step 4: Apply Energy Balance to Solve for Specific Heat of Sample

Using the energy balance equation \( q_{sample} = -(q_{water} + q_{copper}) \), substitute the values we have calculated to solve for the specific heat of the sample. Let \( c_{sample} \) be its specific heat. \( q_{sample} = m_{sample} \cdot c_{sample} \cdot (T_{final} - T_{initial, sample}) \) \[ q_{sample} = 85 \cdot c_{sample} \cdot (26.1 - 100.0) \]\[ -(85 \cdot c_{sample} \cdot -73.9) = 5942 + 410.175 \]\[ 85 \cdot c_{sample} \cdot 73.9 = 6352.175 \]Solving for \( c_{sample} \):\[ c_{sample} = \frac{6352.175}{85 \times 73.9} \approx 1.014 \text{ J/g°C} \]

Step 5: Conclusion

The specific heat capacity of the sample is approximately \( 1.014 \text{ J/g°C} \). This specific heat value helps identify the unknown material by comparing it to known specific heat values of different substances.

Key Concepts

Specific Heat Capacity

Specific heat capacity is a property that indicates how much energy a substance must absorb or release to change its temperature by one degree Celsius per unit mass. It is an intrinsic quality, distinct to each material. This makes it a handy way to identify unknown materials in experiments. In mathematical terms, the specific heat capacity (\( c \)) is often included in the formula:

• \( q = m \cdot c \cdot \Delta T \)

Where:

• \( q \) is the heat energy transferred
• \( m \) is the mass
• \( \Delta T \) is the change in temperature

By knowing the specific heat capacity, it's possible to understand how substances will react to heat conditions, which is essential in many areas, such as climate science, cooking, and engineering.

Energy Balance Equation

The energy balance equation is integral in calorimetry, ensuring that the total energy in a closed system remains constant. In simple terms, it means that energy lost by one part of the system must be gained by another. This is often expressed as:

• \( q_{sample} + q_{water} + q_{copper} = 0 \)

In our specific problem, since there is no heat loss to the surroundings, the sum of the heat exchanges must equal zero. This principle allows us to dissect which parts are gaining or losing heat. Using this concept, you can figure out unknown variables, like the specific heat capacity of a mystery material, by examining known changes throughout the system.

Heat Transfer

Heat transfer is the movement of thermal energy from one object or material to another, driven by a difference in temperature. There are three primary methods of heat transfer: conduction, convection, and radiation. However, in calorimetry, we primarily focus on conduction, where heat transfers directly between substances in contact. In the exercise, heat is transferred from the initially hot sample material to the cooler water and copper can around it until a thermal equilibrium is reached—a point where all temperatures settle at the same value. Understanding heat transfer is crucial for predicting how substances interact when mixed, helping ensure processes like cooking or chemical reactions can be carried out efficiently and safely.

Experimental Physics

Experimental physics is a branch that investigates physical phenomena through experimentation and observation. In calorimetry, an experimental setup is crucial for determining properties like specific heat capacity. The process typically involves precise measurement of masses, the specific heat capacities of reference materials like water and copper, and accurate temperature readings. Key aspects include ensuring minimal heat loss to the surroundings and correct interpretation of data. Through experimental physics, the theoretical principles are put to the test, informing everything from scientific research to practical applications in engineering and materials science. It involves a lot of meticulous work but provides the tangible results necessary for scientific advancement.