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Magazine Chapter 14: Problem 20
(II) A 215-g sample of a substance is heated to 330\(^\circ\)C and then plunged into a 105-g aluminum calorimeter cup containing 185 g of water and a 17-g glass thermometer at 10.5\(^\circ\)C. The final temperature is 35.0\(^\circ\)C. What is the specific heat of the substance? (Assume no water boils away.)
Short Answer
Expert verified
The specific heat of the substance is approximately 0.49 J/g掳C.
Step by step solution
01
Understand the Problem
We need to determine the specific heat of the unknown substance when it is mixed with water, an aluminum cup, and a glass thermometer. We know the initial and final temperatures and the masses of each component involved.
02
Identify Known Values
Here are the given values:- Mass of the substance, \( m_s = 215 \, \text{g} \)- Initial temperature of the substance, \( T_{si} = 330 \degree C \)- Mass of the aluminum cup, \( m_a = 105 \, \text{g} \)- Mass of the water, \( m_w = 185 \, \text{g} \)- Mass of the glass thermometer, \( m_g = 17 \, \text{g} \)- Initial temperature of the cup, water, and thermometer, \( T_{wi} = 10.5 \degree C \)- Final temperature for all components, \( T_f = 35.0 \degree C \).
03
Write the Heat Transfer Equations
The heat lost by the substance = heat gained by water + heat gained by the aluminum cup + heat gained by the thermometer. Using the specific heat formula, we write:- \( q_s = m_s \cdot c_s \cdot (T_f - T_{si}) \) - \( q_w = m_w \cdot c_w \cdot (T_f - T_{wi}) \) - \( q_a = m_a \cdot c_a \cdot (T_f - T_{wi}) \)- \( q_g = m_g \cdot c_g \cdot (T_f - T_{wi}) \)where \( c_s \) is the specific heat of the substance we want to find, \( c_w = 4.18 \, \text{J/g}\degree C \) for water, \( c_a = 0.897 \, \text{J/g}\degree C \) for aluminum, and \( c_g = 0.84 \, \text{J/g}\degree C \) for glass.
04
Set Up the Heat Balance Equation
The heat lost by the substance equals the sum of the heats gained by the other components:\[ m_s \cdot c_s \cdot (T_f - T_{si}) = m_w \cdot c_w \cdot (T_f - T_{wi}) + m_a \cdot c_a \cdot (T_f - T_{wi}) + m_g \cdot c_g \cdot (T_f - T_{wi}). \]
05
Calculate Each Component's Heat Change
Calculate each term in the equation:- Water: \( q_w = 185 \, \text{g} \cdot 4.18 \, \text{J/g}\degree C \cdot (35.0 - 10.5) \degree C \)- Aluminum: \( q_a = 105 \, \text{g} \cdot 0.897 \, \text{J/g}\degree C \cdot (35.0 - 10.5) \degree C \)- Glass: \( q_g = 17 \, \text{g} \cdot 0.84 \, \text{J/g}\degree C \cdot (35.0 - 10.5) \degree C \)- Substance change: \( m_s \cdot c_s \cdot (35.0 - 330) \degree C \) (Note: \( T_{si} > T_f \), hence negative heat loss).
06
Solve for the Specific Heat of the Substance
Simplify and solve the heat balance equation for \( c_s \):\[ 215 \, \text{g} \cdot c_s \cdot (35.0 - 330) \degree C = \sum q_{gain} \] Solve for \( c_s \).
07
Compute to Find Specific Heat
Using calculated values: \( \sum q_{gain} = q_w + q_a + q_g \) \( -215 \, \text{g} \cdot c_s \cdot 295 = \sum q_{gain} \)\( c_s = \frac{\sum q_{gain}}{-215 \cdot -295} \).
08
Final Calculation
Calculate each term's energy and add them, compute the final equation. The specific heat (\(c_s\)) of the substance gives us a positive numerical value in \( \text{J/g}\degree C \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Heat Transfer
Whenever two objects at different temperatures come into contact, heat will flow from the hotter object to the cooler one until they reach the same temperature. This process is known as heat transfer. In our exercise, the hot substance transfers its heat to the cooler aluminum calorimeter cup, water, and glass thermometer.
Heat transfer is driven by the temperature difference between the two objects. The greater the difference, the faster the transfer occurs, initially. It's important to know that heat always moves from a warmer environment to a cooler one, trying to achieve equilibrium. The total amount of heat transfer can be calculated if we know the specific heats and masses of the materials involved, alongside their temperature changes.
Heat transfer is driven by the temperature difference between the two objects. The greater the difference, the faster the transfer occurs, initially. It's important to know that heat always moves from a warmer environment to a cooler one, trying to achieve equilibrium. The total amount of heat transfer can be calculated if we know the specific heats and masses of the materials involved, alongside their temperature changes.
- The formula used for calculating the heat transfer is given by \( q = m \cdot c \cdot \Delta T \). Here, \( q \) is the heat transferred, \( m \) represents the mass, \( c \) is the specific heat of the substance, and \( \Delta T \) is the change in temperature from start to finish.
- The reason specific heat is important is that it tells us how much energy is needed to change the temperature of one gram of a substance by one degree Celsius.
Calorimetry
Calorimetry is the science of measuring heat transfer during physical or chemical changes. This technique uses a calorimeter, an insulated device that helps measure changes in energy. In our exercise, an aluminum calorimeter cup is used. Calorimetry allows us to understand and calculate the specific heat of a substance by monitoring how it affects the temperature of its surroundings.
Calorimetry involves a detailed setup where the heat lost by a hot object is equal to the heat gained by the other materials, including the cup, thermometer, and water. It allows for an understanding of the specific heat by accurately calculating heat transfer. This way, you can determine unknown specific heats of materials as demonstrated in this exercise.
Calorimetry involves a detailed setup where the heat lost by a hot object is equal to the heat gained by the other materials, including the cup, thermometer, and water. It allows for an understanding of the specific heat by accurately calculating heat transfer. This way, you can determine unknown specific heats of materials as demonstrated in this exercise.
- The main idea is the conservation of energy, which indicates that energy cannot be created or destroyed in an isolated system.
- In calorimetry, the setup is built such that the only heat exchange happens within the components of the calorimeter.
Thermal Equilibrium
Thermal equilibrium is the state reached when two or more bodies exchange heat energy until they all reach the same temperature. In the problem we are looking at, the substance, aluminum cup, water, and thermometer eventually reach a final, common temperature of 35.0\(^\circ \)C.
Reaching thermal equilibrium means there is no net heat flow between the bodies involved. All parts involved in the calorimeter setup stop exchanging heat when equilibrium is achieved. This is because the total heat lost by the warmer objects equals the total heat gained by the cooler ones.
Reaching thermal equilibrium means there is no net heat flow between the bodies involved. All parts involved in the calorimeter setup stop exchanging heat when equilibrium is achieved. This is because the total heat lost by the warmer objects equals the total heat gained by the cooler ones.
- This concept is important for understanding practical applications of heat transfer and calorimetry, as it defines the completion of the thermal interaction.
- In real experiments, reaching thermal equilibrium is essential for accurate calorimetry measurements, as this goal ensures thorough heat exchange within the calorimeter.
