Intro to Calorimetry: Videos & Practice Problems

Calorimetry involves calculating the final equilibrium temperature when two materials at different temperatures mix. The heat gained by the colder material equals the heat lost by the hotter one, expressed as $Q$_A=-Q_B. For multiple materials, the general equation is $T$_final=m_Ac_AT_A+m_Bc_BT_B+m_Ac_A+m_Bc_B. Understanding these principles is crucial for solving thermal equilibrium problems effectively.

Solving Calorimetry Problems

In calorimetry, we analyze the heat transfer between two or more substances at different temperatures until they reach thermal equilibrium, where no further heat transfer occurs. The fundamental principle is that the heat lost by the hotter substance equals the heat gained by the colder substance, expressed mathematically as:

$$Q_A = -Q_B$$

To solve calorimetry problems, we often use the equation:

$$Q = mc\Delta T$$

where:

• Q is the heat absorbed or released (in joules),
• m is the mass of the substance (in kilograms),
• c is the specific heat capacity (for water, approximately 4.18 J/g°C), and
• ΔT is the change in temperature (final temperature - initial temperature).

In a typical problem, consider two quantities of water: 1 kg at 20°C and 5 kg at 90°C. When mixed, they will reach a final equilibrium temperature (T_final) that lies between the two initial temperatures. The steps to find T_final are as follows:

1. Set up the equation: $$m_A c \Delta T_A = -m_B c \Delta T_B$$
2. Substitute the specific heat of water (c) and simplify, as it cancels out on both sides.
3. Express the temperature changes: $$\Delta T_A = T_{final} - T_{A_{initial}}$$ and $$\Delta T_B = T_{final} - T_{B_{initial}}$$.
4. Plug in the known values and solve for T_final.

For our example, substituting the values gives:

$$1 \cdot (T_{final} - 20) = -5 \cdot (T_{final} - 90)$$

Expanding and rearranging leads to:

$$T_{final} = \frac{470}{6} \approx 78.3°C$$

This result indicates that the final temperature is closer to the initial temperature of the larger mass of water (90°C), demonstrating how the mass and temperature influence the equilibrium temperature, similar to the concept of center of mass in physics.

Mixing Hot & Cold Water

In this calorimetry problem, we are tasked with finding the mass of boiling water needed to achieve a final temperature of 80 degrees Celsius when mixed with 0.5 kilograms of water at 15 degrees Celsius. The principle behind this is based on the conservation of energy, where the heat gained by the cooler water equals the heat lost by the boiling water.

To start, we denote the heat gained by the cooler water (Q_A) as equal to the negative of the heat lost by the boiling water (Q_B), expressed mathematically as:

Q_A = -Q_B

Next, we apply the formula for heat transfer, which is:

Q = mcΔT

Here, m represents mass, c is the specific heat capacity (for water, c = 4186 J/kg·°C), and ΔT is the change in temperature. For the cooler water (A), we have:

Q_A = m_AcΔT_A = (0.5 kg)(4186 J/kg·°C)(80°C - 15°C)

For the boiling water (B), we express its heat loss as:

Q_B = m_BcΔT_B = m_B(4186 J/kg·°C)(80°C - 100°C)

Since both sides of the equation involve the same specific heat capacity, we can simplify our calculations by canceling it out:

0.5 kg × (80°C - 15°C) = -m_B × (80°C - 100°C)

Calculating the temperature changes, we find:

ΔT_A = 80°C - 15°C = 65°C

ΔT_B = 80°C - 100°C = -20°C

Substituting these values into the equation gives:

0.5 kg × 65°C = -m_B × (-20°C)

Solving for m_B leads to:

32.5 = 20m_B

m_B = 1.625 kg

This result indicates that 1.625 kilograms of boiling water is required to achieve the desired final temperature of 80 degrees Celsius. The larger mass of boiling water compared to the cooler water is consistent with the final temperature being closer to the boiling point, reflecting the greater thermal energy content of the boiling water.

Equilibrium Temperature Equation in Calorimetry Problems

In calorimetry, determining the equilibrium temperature of a mixture of materials is a fundamental process. This involves understanding how heat transfers between substances until thermal equilibrium is reached. The key principle is that the heat lost by the hotter substance equals the heat gained by the colder substance, which can be expressed mathematically as \( Q_A = -Q_B \).

To derive a general equation for the final equilibrium temperature, we start by substituting the heat transfer equations for each material. The heat transfer for each substance can be represented as \( Q = mc\Delta T \), where \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. For two materials, this leads to the equation:

\( m_A c_A (T_{final} - T_A) = -m_B c_B (T_{final} - T_B) \)

Here, \( T_A \) is the initial temperature of the colder substance (e.g., water), and \( T_B \) is the initial temperature of the hotter substance (e.g., aluminum). By expanding and rearranging the equation, we can isolate \( T_{final} \) to find:

\( T_{final} = \frac{m_A c_A T_A + m_B c_B T_B}{m_A c_A + m_B c_B} \

This formula can be generalized for any number of materials. For \( n \) materials, the equilibrium temperature can be expressed as:

\( T_{final} = \frac{\sum_{i=1}^{n} m_i c_i T_i}{\sum_{i=1}^{n} m_i c_i} \

In practical applications, such as calculating the equilibrium temperature of a mixture of water and aluminum, specific heat values are essential. For water, the specific heat capacity \( c_A \) is approximately 4186 J/(kg·°C), while for aluminum, \( c_B \) is about 900 J/(kg·°C). By substituting the known values into the derived equation, we can compute the final equilibrium temperature.

For example, if we have 0.4 kg of water at 10 °C and 0.2 kg of aluminum at 80 °C, we can plug these values into the equation:

\( T_{final} = \frac{(0.4 \, \text{kg} \times 4186 \, \text{J/(kg·°C)} \times 10 \, °C) + (0.2 \, \text{kg} \times 900 \, \text{J/(kg·°C)} \times 80 \, °C)}{(0.4 \, \text{kg} \times 4186 \, \text{J/(kg·°C)}) + (0.2 \, \text{kg} \times 900 \, \text{J/(kg·°C)})} \

After performing the calculations, the final equilibrium temperature is found to be approximately 16.7 °C. This process illustrates the application of calorimetry principles in determining the thermal behavior of mixed substances.

Pouring Hot Water in an Aluminum Cup

In this calorimetry problem, we are tasked with finding the final equilibrium temperature when mixing two substances: 150 grams of water at 35°C and a 65-gram aluminum cup at 11°C. The principle of conservation of energy applies here, where the heat lost by the water equals the heat gained by the aluminum cup. This relationship can be expressed with the equation:

Q_aluminum = -Q_water

To find the final temperature (T_final), we can use the equilibrium temperature equation, which is structured as follows:

T_final = \frac{(m_aluminum \cdot c_aluminum \cdot T_aluminum) + (m_water \cdot c_water \cdot T_water)}{(m_aluminum \cdot c_aluminum) + (m_water \cdot c_water)}

Here, m represents mass, c represents specific heat capacity, and T represents temperature. For aluminum, the specific heat capacity (c_aluminum) is 910 J/(kg·°C), while for water (c_water), it is 4186 J/(kg·°C).

Before plugging in the values, it is essential to convert the masses from grams to kilograms:

• m_aluminum = 65 g = 0.065 kg
• m_water = 150 g = 0.15 kg

Now, substituting the values into the equation:

T_final = \frac{(0.065 \cdot 910 \cdot 11) + (0.15 \cdot 4186 \cdot 35)}{(0.065 \cdot 910) + (0.15 \cdot 4186)}

Calculating the numerator and denominator separately, we find:

Numerator: (0.065 × 910 × 11) + (0.15 × 4186 × 35) = 0.065 × 910 × 11 + 0.15 × 4186 × 35

Denominator: (0.065 × 910) + (0.15 × 4186) = 0.065 × 910 + 0.15 × 4186

After performing the calculations, we arrive at a final temperature of approximately 32.9°C. This result is logical, as it lies between the initial temperatures of 11°C and 35°C, and is closer to 35°C due to the larger mass and higher specific heat capacity of water compared to aluminum.

Here’s what students ask on this topic:

What is the principle of calorimetry?

The principle of calorimetry is based on the law of conservation of energy, which states that energy cannot be created or destroyed, only transferred. In calorimetry, this principle is applied to heat transfer between substances. When two materials at different temperatures are mixed, the heat lost by the hotter material is equal to the heat gained by the colder material. This can be expressed as:

$Q{A}_{}=-Q_{}{B}_{}$

where $Q$ represents the heat transfer. This principle allows us to calculate the final equilibrium temperature of the mixture.

How do you solve calorimetry problems involving two materials?

To solve calorimetry problems involving two materials, follow these steps:

1. Write the equation for heat transfer: $Q_A=-Q_B$.

2. Replace $Q$ with $\mathrm{mc\Delta T}$ for both materials.

3. Expand the temperature change terms: $\mathrm{\Delta T}$ is $T$_final-T_initial.

4. Simplify and solve for the final equilibrium temperature $T$_final.

This process ensures that the heat lost by the hotter material equals the heat gained by the colder material, leading to thermal equilibrium.

What is the general equation for the final equilibrium temperature in calorimetry?

The general equation for the final equilibrium temperature $T$_final in calorimetry, when mixing multiple materials, is:

$T$_final=mAcATA+mBcBTB+...mAcA+mBcB+...

This equation accounts for the masses, specific heats, and initial temperatures of all materials involved. It ensures that the heat gained by the colder materials equals the heat lost by the hotter materials, leading to thermal equilibrium.

How does the specific heat capacity affect the final equilibrium temperature in calorimetry?

The specific heat capacity of a material affects the final equilibrium temperature in calorimetry by determining how much heat is required to change the temperature of the material. A higher specific heat capacity means the material can absorb more heat without a significant change in temperature. In the equation:

$T$_final=mAcATA+mBcBTB+...mAcA+mBcB+...

the specific heat capacities $c$ influence the final temperature. Materials with higher specific heat capacities will have a smaller temperature change, affecting the final equilibrium temperature.

What is thermal equilibrium in the context of calorimetry?

In the context of calorimetry, thermal equilibrium is the state reached when two or more materials at different temperatures are mixed, and their temperatures become equal. At this point, there is no net heat transfer between the materials. The heat lost by the hotter material equals the heat gained by the colder material, leading to a stable final temperature. This concept is crucial for solving calorimetry problems, as it allows us to calculate the final equilibrium temperature using the principle of conservation of energy.