Problem 56 - University Physics with Modern Physics

Question

A laboratory technician drops a \(0.0850-\mathrm{kg}\) sample of unknown solid material, at a temperature of \(100.0^{\circ} \mathrm{C},\) into a calorimeter. The calorimeter can, initially at \(19.0^{\circ} \mathrm{C},\) is made of 0.150 \(\mathrm{kg}\) of copper and contains 0.200 \(\mathrm{kg}\) of water. The final temperature of the calorimeter can and contents is \(26.1^{\circ} \mathrm{C}\) . Compute the specific heat of the sample.

Step-by-Step Solution

Answer

The specific heat of the sample is calculated approximately as 0.724 J/g°C.

Step 1: Identify the Known Values

We are given that the mass of the sample is \( m_s = 0.0850 \) kg, the initial temperature of the sample \( T_{s_i} = 100.0^{\circ} \mathrm{C} \), and the final temperature of the system \( T_f = 26.1^{\circ} \mathrm{C} \). For the calorimeter, the mass of the copper can \( m_c = 0.150 \) kg, the specific heat of copper \( c_c = 0.385 \) J/g°C, and the initial temperature \( T_{c_i} = 19.0^{\circ} \mathrm{C} \). The water has a mass \( m_w = 0.200 \) kg, the specific heat \( c_w = 4.186 \) J/g°C, and the same initial temperature as the calorimeter.

Step 2: Calculate Heat Gained by Water

Calculate the heat gained by the water using the formula \( q_w = m_w \cdot c_w \cdot (T_f - T_{w_i}) \), where \( T_{w_i} = 19.0^{\circ} \mathrm{C} \). Thus, \( q_w = 200 \times 4.186 \times (26.1 - 19.0) \).

Step 3: Calculate Heat Gained by Copper

Calculate the heat gained by the copper using the formula \( q_c = m_c \cdot c_c \cdot (T_f - T_{c_i}) \). Therefore, \( q_c = 150 \times 0.385 \times (26.1 - 19.0) \).

Step 4: Sum of Heat Gained by Water and Copper

Calculate the total heat gained by adding the heat gained by the water and the heat gained by the copper. This is given by \( q_{total} = q_w + q_c \).

Step 5: Calculate Heat Lost by Sample

Using the principle of conservation of energy, the heat lost by the sample is equal to the total heat gained by the calorimeter system: \( q_s = -q_{total} \).

Step 6: Calculate Specific Heat of the Sample

The heat lost by the sample is calculated as \( q_s = m_s \cdot c_s \cdot (T_f - T_{s_i}) \). Solve for \( c_s \) using \( c_s = \frac{q_s}{m_s \cdot (T_f - T_{s_i})} \).

Step 7: Plug in Values and Solve

Substitute the known values into the equation for \( c_s \) from Step 6, using the value of \( q_{total} \) calculated in Step 4. Compute the specific heat \( c_s \) for the sample.

Key Concepts

Calorimetry

Calorimetry is a fascinating branch of science that deals with the measurement of heat transfer during chemical reactions or physical changes. In a calorimetry experiment, a calorimeter is used to detect and quantify this heat change. The calorimeter functions as an insulated container to minimize heat loss to the environment. This helps ensure that the heat transfer observed is mainly between the substances in the calorimeter.

The process typically involves a substance at a certain temperature being introduced into the calorimeter, where it causes a temperature change in the enclosed fluids, such as water. We measure the temperature difference to quantify the amount of energy exchanged.

In our exercise, the unknown solid material was placed into a calorimeter containing water and a copper can. The temperature changed from 19.0°C to 26.1°C. By analyzing these temperature shifts, we can deduce the specific heat of the unknown solid material.

Conservation of Energy

The principle of conservation of energy is a fundamental concept in physics. It asserts that energy cannot be created or destroyed; it can only change forms. In the context of a calorimetry experiment, energy conservation is at the heart of understanding how heat exchange between objects works.

When the solid sample was introduced to the calorimeter, the system exchanged heat. The solid cooled down, releasing its energy, while the water and copper can absorbed this energy. According to the conservation of energy, the heat lost by the sample should equal the total heat gained by the calorimeter contents: water and copper.

Mathematically, we express this as:

• Heat lost by the sample ( \[ q_s \] ) = heat gained by the calorimeter contents
• - \[ q_{sample} \] = \[ q_w + q_c \]

This key equation allows us to solve for unknowns, such as the specific heat capacity of a material.

Heat Transfer

Heat transfer is the process through which thermal energy moves from one body to another as a result of temperature difference. In our calorimetry scenario, heat transfer is critical, as it determines how energy flows between the material sample, water, and the copper can.

The transfer of heat in calorimetry is predominantly governed by specific heat capacity, a property that describes how much energy is required to raise the temperature of a unit mass of a substance by one degree Celsius. Materials with high specific heat capacity can absorb more heat without experiencing a large change in temperature.

In the experiment, we assessed the heat gained by both the water and copper:

• Heat gained by water: \[ q_w = m_w \cdot c_w \cdot (T_f - T_{w_i}) \]
• Heat gained by copper can: \[ q_c = m_c \cdot c_c \cdot (T_f - T_{c_i}) \]

These calculations help us paint the complete picture of energy transfer within the calorimeter system. In turn, they enable us to determine how much heat the unknown material loses and, consequently, to measure its specific heat capacity.