Question

A 155-g sample of an unknown substance was heated from 25.0°C to 40.0°C. In the process, the substance absorbed 5696 J of energy. What is the specific heat of the substance? Identify the substance among those listed in Table 15.2.

Short answer

The specific heat of the unknown substance is approximately 2.45 J/(g°C). By comparing this value to the specific heat values in Table 15.2, we can identify the unknown substance. For example, if the table indicates copper has a specific heat close to 2.45 J/(g°C), we can conclude that the unknown substance is likely copper.

Step-by-step solution

1. Calculate the change in temperature

First, we should calculate the change in temperature (ΔT). To calculate the change in temperature, we subtract the initial temperature from the final temperature: ΔT = T_final - T_initial. ΔT = 40.0°C - 25.0°C ΔT = 15.0°C

2. Rearrange the formula to solve for specific heat (c)

We have the formula Q = mcΔT, and we want to find the specific heat (c). To solve for c, we will rearrange the formula by dividing both sides of the equation by mΔT: c = Q / (mΔT) We already know the values for Q, m, and ΔT, so we can plug them into the formula and calculate the specific heat.

3. Calculate the specific heat (c)

Now, we can plug in the values we have (Q = 5696 J, m = 155 g, and ΔT = 15.0°C) into the formula: c = 5696 J / (155 g * 15.0°C) c = 5696 J / 2325 g°C c ≈ 2.45 J/(g°C) The specific heat of the unknown substance is approximately 2.45 J/(g°C).

4. Identify the substance

Now that we have the specific heat value of the substance, we can use the information from Table 15.2 (not provided here) to identify the substance. By comparing the specific heat value we found (2.45 J/(g°C)) to the values in the table, we can determine which substance most closely matches our calculated specific heat. For example, if the table lists the specific heat of water as 4.18 J/(g°C), aluminum as 0.897 J/(g°C), and another substance like copper has a value close to 2.45 J/(g°C), we can conclude that the unknown substance is likely to be copper (assuming copper is listed in Table 15.2).

Key concepts

Temperature Change

When dealing with temperature changes in a substance, it's essential to understand how it impacts the material's heat energy absorption or release. The temperature change, denoted as \( \Delta T \), helps us quantify this shift. To calculate \( \Delta T \), subtract the initial temperature from the final temperature:

• Final Temperature (\( T_{\text{final}} \))
• Initial Temperature (\( T_{\text{initial}} \))

Thus, \( \Delta T = T_{\text{final}} - T_{\text{initial}} \). In the example, the substance starts at 25.0°C and ends at 40.0°C.By performing the subtraction, we find \( \Delta T = 15.0\,^{\circ}\text{C} \). This temperature increase shows how much the substance's temperature rose due to the absorbed heat energy.

Heat Energy

Heat energy, denoted as \( Q \), is a fascinating aspect of thermal dynamics. It represents the energy transferred due to temperature differences. When a substance absorbs or releases heat, it undergoes a temperature change related to the material's mass and specific heat capacity.For our calculations, given heat energy \( Q = 5696\,\text{J} \), mass \( m = 155\,\text{g} \), and temperature change \( \Delta T = 15.0\,^{\circ}\text{C} \), the equation is:\[ Q = mc\Delta T \]Where:

• \(c\) is the specific heat capacity
• \(m\) is the mass
• \(\Delta T\) is the change in temperature

Rearranging to solve for specific heat capacity, we get:\[ c = \frac{Q}{m \cdot \Delta T} \]By substituting the known values, we calculate the specific heat to be approximately \( 2.45\,\text{J/(g\,^{\circ}\text{C})} \). This value provides insight into how the substance stores and transfers heat energy.

Substance Identification

Identifying a substance based on its specific heat capacity involves comparing calculated values with known data. The specific heat capacity is unique to each material, making it an excellent identifier. Once you have calculated the specific heat, check against a reference table (ideally provided in lab or textbook resources).

• For instance, water typically has a specific heat of \( 4.18\,\text{J/(g\,^{\circ}\text{C})} \)
• Aluminum is around \( 0.897\,\text{J/(g\,^{\circ}\text{C})} \)
• Copper could be close to our example calculation of \( 2.45\,\text{J/(g\,^{\circ}\text{C})} \)

In our scenario, assuming copper is listed in the reference table with a specific heat similar to \( 2.45\,\text{J/(g\,^{\circ}\text{C})} \), the unknown substance could likely be copper. This identification process highlights the role of specific heat as a fingerprint for materials in thermal circumstances.