Problem 1
Question
Heat Effects and Calorimetry A metal sample weighing \(147.90 \mathrm{g}\) and at a temperature of \(99.5^{\circ} \mathrm{C}\) was placed in \(49.73 \mathrm{g}\) of water in a calorimeter at \(23.0^{\circ} \mathrm{C}\). At equilibrium the temperature of the water and metal was \(41.8^{\circ} \mathrm{C}\). a. What was \(\Delta t\) for the water? \(\left(\Delta t=t_{\text {final }}-t_{\text {intilial }}\right)\) ________ \(^{\circ} \mathrm{C}\) b. What was \(\Delta t\) for the metal? ________\(^{\circ} \mathrm{C}\) c. How much heat flowed into the water? (Take the specific heat of the water to be 4.18 J/g \(^{\circ}\) C.) _______ joules d. Calculate the specific heat of the metal, using Equation 3. ______joules/g"C e. What is the approximate molar mass of the metal? (Use Eq. 4.) ______\(g /m o l\)
Step-by-Step Solution
Verified Answer
a. 18.8°C b. -57.7°C c. 3919.56 J d. 0.467 J/g°C e. 55.7 g/mol
1Step 1: Calculate Δt for the water
First, we calculate the temperature change for the water. We use the formula: \[ \Delta t_{ ext{water}} = t_{ ext{final}} - t_{ ext{initial}} \] Substitute the given values: \[ \Delta t_{ ext{water}} = 41.8^{\circ} \mathrm{C} - 23.0^{\circ} \mathrm{C} = 18.8^{\circ} \mathrm{C} \]
2Step 2: Calculate Δt for the metal
Next, we find the temperature change for the metal using a similar approach: \[ \Delta t_{ ext{metal}} = t_{ ext{final}} - t_{ ext{initial}} \] Substitute the given values: \[ \Delta t_{ ext{metal}} = 41.8^{\circ} \mathrm{C} - 99.5^{\circ} \mathrm{C} = -57.7^{\circ} \mathrm{C} \] This means the metal cooled down.
3Step 3: Calculate the heat absorbed by the water
To find the heat absorbed by the water, use the formula: \[ q = mc\Delta t \] where \( m \) is the mass, \( c \) is the specific heat, and \( \Delta t \) is the temperature change.For water: \( m = 49.73 \; \mathrm{g}, c = 4.18 \; \mathrm{J/g^{\circ}C}, \Delta t = 18.8^{\circ} \mathrm{C} \).Plug these values into the equation: \[ q = 49.73 \; \mathrm{g} \times 4.18 \; \mathrm{J/g^{\circ}C} \times 18.8^{\circ} \mathrm{C} \approx 3919.56 \; \mathrm{J} \] The water absorbed approximately 3919.56 joules.
4Step 4: Calculate the specific heat of the metal
Use the formula for heat transfer, considering the heat lost by the metal equals the heat gained by the water:\[ q_{ ext{metal}} = -q_{ ext{water}} \]The formula for metal becomes: \[ q_{ ext{metal}} = m_{ ext{metal}} c_{ ext{metal}} \Delta t_{ ext{metal}} \]Rearranging to solve for \(c_{ ext{metal}}\):\[ c_{ ext{metal}} = \frac{-q_{ ext{water}}}{m_{ ext{metal}} \Delta t_{ ext{metal}}} \]Substitute the values: \[ m_{ ext{metal}} = 147.90 \; \mathrm{g}, \Delta t_{ ext{metal}} = -57.7^{\circ} \mathrm{C}, q_{ ext{water}} = 3919.56 \; \mathrm{J} \]\[ c_{ ext{metal}} = \frac{-3919.56 \; \mathrm{J}}{147.90 \; \mathrm{g} \times -57.7^{\circ} \mathrm{C}} \approx 0.467 \; \mathrm{J/g^{\circ}C} \]
5Step 5: Calculate the molar mass of the metal
Use the formula relating specific heat to molar mass: \[ M = \frac{constant}{c} \] A typical constant for solid metals is approximately \(26 \; \mathrm{J/mol\cdot^{\circ}C}\).Substitute \( c = 0.467 \; \mathrm{J/g^{\circ}C} \):\[ M = \frac{26 \; \mathrm{J/mol\cdot^{\circ}C}}{0.467 \; \mathrm{J/g^{\circ}C}} \approx 55.7 \; \mathrm{g/mol} \] The molar mass of the metal is approximately 55.7 g/mol.
Key Concepts
Heat TransferSpecific Heat CapacityMolar Mass Calculation
Heat Transfer
Heat transfer is a crucial concept in calorimetry. It is the movement of heat energy from one substance to another. In the context of the problem, when a hot metal sample is placed into cooler water, heat moves from the metal to the water. This process continues until both the metal and the water reach a common temperature, known as equilibrium. Understanding how heat transfer works helps us measure changes in temperature, indicating how much heat energy was exchanged between substances.
Heat transfer can occur through three main processes: conduction, convection, and radiation. In this exercise, conduction is at play. Conduction involves the transfer of heat through direct contact between particles of the substances. As the metal touches the water, kinetic energy from the hotter metal molecules transfers to the cooler water molecules, leading to a change in temperature for both.
The amount of heat transferred can be calculated using the formula: \[ q = mc\Delta t \] where \( q \) represents heat transferred, \( m \) is the mass of the substance, \( c \) is the specific heat capacity, and \( \Delta t \) is the change in temperature. This formula is key for quantifying heat transfer in calorimetry experiments.
Heat transfer can occur through three main processes: conduction, convection, and radiation. In this exercise, conduction is at play. Conduction involves the transfer of heat through direct contact between particles of the substances. As the metal touches the water, kinetic energy from the hotter metal molecules transfers to the cooler water molecules, leading to a change in temperature for both.
The amount of heat transferred can be calculated using the formula: \[ q = mc\Delta t \] where \( q \) represents heat transferred, \( m \) is the mass of the substance, \( c \) is the specific heat capacity, and \( \Delta t \) is the change in temperature. This formula is key for quantifying heat transfer in calorimetry experiments.
Specific Heat Capacity
Specific heat capacity defines how much heat energy is required to raise the temperature of a unit mass of a substance by one degree Celsius. It is an intrinsic property of materials. In the exercise, you use the specific heat capacity of water, which is known to be \(4.18 \; \mathrm{J/g^{\circ}C}\), to determine the heat absorbed by water.
The specific heat capacity tells us how a substance will respond to the addition or removal of heat. For example, substances with high specific heat capacities, like water, can absorb or release large amounts of heat with little change in temperature. In constrast, metals usually have lower specific heat capacities, meaning their temperatures shift significantly with relatively small energy exchanges.
In the formula \( q = mc\Delta t \), the specific heat capacity \( c \) helps calculate how much heat (\( q \)) is needed for a specific temperature change (\( \Delta t \)) of a given mass (\( m \)). Manipulating this formula allows you to solve for the specific heat of unknown substances, explaining how different materials react to thermal changes.
The specific heat capacity tells us how a substance will respond to the addition or removal of heat. For example, substances with high specific heat capacities, like water, can absorb or release large amounts of heat with little change in temperature. In constrast, metals usually have lower specific heat capacities, meaning their temperatures shift significantly with relatively small energy exchanges.
In the formula \( q = mc\Delta t \), the specific heat capacity \( c \) helps calculate how much heat (\( q \)) is needed for a specific temperature change (\( \Delta t \)) of a given mass (\( m \)). Manipulating this formula allows you to solve for the specific heat of unknown substances, explaining how different materials react to thermal changes.
Molar Mass Calculation
Molar mass calculation often comes into play when identifying a substance based on its specific heat capacity. By experimentally determining a metal's specific heat capacity, you can infer its molar mass using empirical formulas.
The formula used in the exercise is: \[ M = \frac{\text{constant}}{c} \] where \( M \) is the molar mass, \( c \) is the specific heat capacity, and the constant typically used for metals is approximately \(26 \; \mathrm{J/mol\cdot^{\circ}C}\). This empirical relationship works because different elements have characteristic ratios of molar increase to specific heat.
Solving for the molar mass like in the exercise, you input the determined specific heat into the formula to find the molar mass of the metal. This allows one to identify the metal based on its molar mass, as each element has a unique molar mass value. It's a method similar to fingerprinting that links the thermal properties back to the atomic level.
The formula used in the exercise is: \[ M = \frac{\text{constant}}{c} \] where \( M \) is the molar mass, \( c \) is the specific heat capacity, and the constant typically used for metals is approximately \(26 \; \mathrm{J/mol\cdot^{\circ}C}\). This empirical relationship works because different elements have characteristic ratios of molar increase to specific heat.
Solving for the molar mass like in the exercise, you input the determined specific heat into the formula to find the molar mass of the metal. This allows one to identify the metal based on its molar mass, as each element has a unique molar mass value. It's a method similar to fingerprinting that links the thermal properties back to the atomic level.