Problem 60 - Physics

Question

Equal masses of two different liquids have the same temperature of \(25.0^{\circ} \mathrm{C}\). Liquid A has a freezing point of \(-68.0{ }^{\circ} \mathrm{C}\) and a specific heat capacity of \(1850 \mathrm{~J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right) .\) Liquid \(\mathrm{B}\) has a freezing point of \(-96.0^{\circ} \mathrm{C}\) and a specific heat capacity of \(2670 \mathrm{~J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right) .\) The same amount of heat must be removed from each liquid in order to freeze it into a solid at its respective freezing point. Determine the difference \(L_{\mathrm{f}, \mathrm{A}}-L_{\mathrm{f}, \mathrm{B}}\) between the latent heats of fusion for these liquids.

Step-by-Step Solution

Answer

The difference in latent heats, \(L_{f,A} - L_{f,B}\), is 151,320 J/kg.

Step 1: Define the Problem

We are given two liquids, A and B, initially at the same temperature of 25.0°C. We need to calculate the difference in latent heat of fusion, \(L_{f,A} - L_{f,B}\), using the given specific heat capacities and freezing points.

Step 2: Calculate Heat Removed for Liquid A

For liquid A, we need to cool it from 25.0°C to -68.0°C. The amount of heat removed to just reach its freezing point is calculated by the formula: \[ Q_A = m imes c_A imes (T_i - T_f) \]Here, \(m\) is the mass, \(c_A\) is the specific heat capacity of A (1850 J/kg°C), \(T_i = 25.0°C\), and \(T_f = -68.0°C\). The difference in temperature is \(T_i - T_f = 93°C\). Thus,\[ Q_A = m imes 1850 imes 93 \]

Step 3: Calculate Heat Removed for Liquid B

Similarly, for liquid B, the heat removed to reach the freezing point is given by:\[ Q_B = m imes c_B imes (T_i - T_{f,B}) \]Where \(c_B\) is 2670 J/kg°C, and \(T_{f,B} = -96.0°C\). The temperature change is \(25.0 - (-96.0) = 121°C\). Thus,\[ Q_B = m imes 2670 imes 121 \]

Step 4: Establish Equal Heat Removal

Since the same amount of heat is removed from both liquids to reach their freezing points and freeze them:\[ Q_A + m imes L_{f,A} = Q_B + m imes L_{f,B} \]This means:\[ m imes 1850 imes 93 + m imes L_{f,A} = m imes 2670 imes 121 + m imes L_{f,B} \]

Step 5: Solve for the Difference in Latent Heats

Canceling mass \(m\) from the equation...\[ 1850 imes 93 + L_{f,A} = 2670 imes 121 + L_{f,B} \]Rearranging gives:\[ L_{f,A} - L_{f,B} = 2670 imes 121 - 1850 imes 93 \]

Step 6: Calculate Numerical Result

Compute the values:\[ 2670 imes 121 = 323370 \]And,\[ 1850 imes 93 = 172050 \]Thus,\[ L_{f,A} - L_{f,B} = 323370 - 172050 = 151320 \]

Key Concepts

Specific Heat Capacity

Specific heat capacity is a crucial concept when analyzing how substances absorb and release heat. It essentially measures how much energy in joules is needed to raise the temperature of 1 kilogram of a substance by 1°C.

• The unit for specific heat capacity is \( ext{J/kg} \, ^{\circ}\text{C}\).
• Higher specific heat capacities mean the substance can absorb more heat before its temperature rises.
• This property varies between different materials.

In the exercise, we see two liquids, A and B, each with its own specific heat capacity—1850 \(\text{J}/(\text{kg} \cdot ^{\circ}\text{C})\) for liquid A and 2670 \(\text{J}/(\text{kg} \cdot ^{\circ}\text{C})\) for liquid B. This informs us that liquid B requires more energy than A to increase its temperature to the same extent, indicating greater resistance to temperature change.

Freezing Point

The freezing point of a substance is the temperature at which it changes from a liquid to a solid state. This is a significant aspect when studying phase changes and energy transfer.

• Different substances have different freezing points.
• The temperature remains constant during the phase change from liquid to solid.
• Energy removal, necessary to switch states, is represented by the latent heat of fusion.

In the exercise, liquid A has a freezing point of \(-68.0^{\circ}\text{C}\) and liquid B a much lower \(-96.0^{\circ}\text{C}\). These values demonstrate that liquid B solidifies at a colder temperature compared to liquid A, specifying the energy required doesn't just depend on specific heat capacity but also the freezing temperature.

Temperature Change

The concept of temperature change is foundational when discussing thermal energy transfer since it directly affects how much energy is needed to accomplish a certain temperature shift.

• Calculated as the final temperature minus the initial temperature (\(T_f - T_i\)).
• For a cooling process, temperature change is typically a negative value, indicating a drop in temperature.
• Changes dictate how much heat energy must be absorbed or relieved.

The exercise involves two specific temperature changes: \(-68^{\circ}\text{C}-25^{\circ}\text{C} = 93^{\circ}\text{C}\) for liquid A, and \(-96^{\circ}\text{C}-25^{\circ}\text{C} = 121^{\circ}\text{C}\) for liquid B. These changes are crucial to calculate the heat removed, helping understand how different substances respond to cooling under identical initial conditions.