Question

A \(50.75-\mathrm{g}\) sample of water at \(75.6^{\circ} \mathrm{C}\) is added to a sample of water at \(24.1^{\circ} \mathrm{C}\) in a constant-pressure calorimeter. If the final temperature of the combined water is \(39.4^{\circ} \mathrm{C}\) and the heat capacity of the calorimeter is \(26.3 \mathrm{~J} /{ }^{\circ} \mathrm{C},\) calculate the mass of the water originally in the calorimeter.

Short answer

The mass of the water originally in the calorimeter is approximately 126.37 grams.

Step-by-step solution

Identify the known quantities

We know the mass of the first water sample is 50.75 g at an initial temperature of 75.6°C. The second water sample is unknown mass (let's call it \(m_2\)), initially at 24.1°C. The final temperature is 39.4°C and the calorimeter's heat capacity is 26.3 J/°C.

Calculate the heat lost by the first water sample

The specific heat capacity of water is approximately 4.18 J/g°C. The heat lost, \(q_1\), by the first sample is calculated by: \[ q_1 = m_1 c \Delta T_1 = 50.75 \: \mathrm{g} \times 4.18 \: \mathrm{J/g°C} \times (39.4 - 75.6) \: °\mathrm{C} \]This gives: \[ q_1 = 50.75 \times 4.18 \times (-36.2) = -7681.833 \: \mathrm{J} \]The negative sign indicates heat lost.

Calculate the heat gained by the second water sample

The heat gained, \(q_2\), by the unknown mass of water is given by: \[ q_2 = m_2 c \Delta T_2 = m_2 \times 4.18 \: \mathrm{J/g°C} \times (39.4 - 24.1) \: °\mathrm{C} \]This simplifies to: \[ q_2 = m_2 \times 4.18 \times 15.3 = 63.954 \times m_2 \: \mathrm{J} \]

Calculate heat absorbed by the calorimeter

The calorimeter absorbs heat as well, calculated by: \[ q_{\text{cal}} = C_{\text{cal}} \Delta T = 26.3 \: \mathrm{J/°C} \times (39.4 - 24.1) \: °\mathrm{C} \]This gives:\[ q_{\text{cal}} = 26.3 \times 15.3 = 402.39 \: \mathrm{J} \]

Set up the heat balance equation

Since there is no heat loss to the surroundings (law of conservation of energy), the total heat lost by the first water equals the heat gained by the second water and the calorimeter: \[ q_1 = q_2 + q_{\text{cal}} \]Substitute known values:\[ -7681.833 = 63.954 \times m_2 + 402.39 \]

Solve for the unknown mass \(m_2\)

Rearrange the equation to solve for \(m_2\):\[ 63.954 \times m_2 = -7681.833 - 402.39 = -8084.223 \]\[ m_2 = \frac{-8084.223}{63.954} \approx 126.37 \: \mathrm{grams} \]

Conclusion

The mass of the water originally in the calorimeter is approximately 126.37 grams.

Key concepts

Heat Capacity

When working with calorimetry, understanding the concept of heat capacity is essential. Heat capacity refers to the amount of heat energy needed to raise the temperature of a substance by one degree Celsius (°C). It is an extensive property, meaning it is dependent on the mass and size of the material being heated.
Knowing the heat capacity allows us to calculate how much energy is required to change the temperature of the entire system. In the context of calorimetry, it pertains to both the substance being studied and the container housing it. For example, a calorimeter's heat capacity is crucial because it absorbs some heat during the reaction, affecting the measurement results.

Specific Heat Capacity

Specific heat capacity is a more refined term related to heat capacity. It tells us the energy required to raise the temperature of a unit mass of a substance by one degree Celsius. Unlike heat capacity, specific heat capacity is an intensive property, meaning it does not rely on the sample size.
Specific heat capacity is measured in Joules per gram per degree Celsius (J/g°C). For example, water has a specific heat capacity of approximately 4.18 J/g°C. This detail is vital in calorimetry calculations. It allows us to relate the heat energy change to the mass and temperature change of water or any given substance.
When solving problems involving mixing or heat exchange, knowing the specific heat helps us determine how the energy will affect the different masses and temperatures involved.

Conservation of Energy

Conservation of energy is a fundamental principle used in calorimetry. This law states that energy cannot be created or destroyed, only transferred or converted from one form to another. When we consider a calorimetry problem, the energy lost by one substance is gained by another.
In the exercise, the heat lost by the warmer water sample is equal to the heat gained by the cooler water and the calorimeter itself. This balance is crucial for accurately calculating unknown variables, such as the mass of one of the water samples. Using the conservation of energy assures us that all energy changes are accounted for.
This principle simplifies complex systems by allowing us to focus solely on the energy transformations that occur within the system, avoiding any need to consider energy interactions with the surroundings.

Water Temperature

Temperature plays a pivotal role in calorimetry problems, particularly the initial, final, and change in temperature of the substances involved. In our particular exercise, the water samples start at different temperatures and are brought to a common final temperature.
The initial temperatures of the substances determine the direction of heat flow. Heat naturally moves from the warmer substance to the cooler one until thermal equilibrium is reached.
Understanding how temperature changes affect the system helps us use calorimetric equations to find unknown values, such as mass. Temperature differentials are used along with specific heat capacity to determine how much energy flows from one part of the system to another.

• Heat flows from high to low.
• Equilibrium is reached when both substances have the same temperature.
• Final temperature is critical for calculating total energy transfer.