A 0.5 G Sample Of $C_{10}H_8$ Is Burned In A Calorimeter That Contains 650 G Of Water At $20.0^{\circ} C$. The Heat Capacity Of The Calorimeter Is $420 \frac{J}{{ }^{\circ} C}$. If The Final Temperature Of The Water Is

$$

Introduction

In this article, we will delve into the world of chemistry and explore the concept of combustion reactions. Specifically, we will analyze the burning of a 0.5 g sample of $C_{10}H_8$ in a calorimeter containing 650 g of water at $20.0^{\circ} C$. We will also examine the heat capacity of the calorimeter and its impact on the final temperature of the water.

The Combustion Reaction

The combustion reaction of $C_{10}H_8$ can be represented by the following equation:

$$C_{10}H_8 + 12.5O_2 \rightarrow 10CO_2 + 4H_2O$$

In this reaction, $C_{10}H_8$ reacts with oxygen to produce carbon dioxide and water. The heat of combustion of $C_{10}H_8$ is a measure of the energy released during this reaction.

The Calorimeter

A calorimeter is a device used to measure the heat of a chemical reaction. In this case, the calorimeter contains 650 g of water at $20.0^{\circ} C$. The heat capacity of the calorimeter is given as $420 \frac{J}{{ }^{\circ} C}$.

The Heat of Combustion

The heat of combustion of $C_{10}H_8$ can be calculated using the following equation:

$$Q = mc\Delta T$$

where $Q$ is the heat of combustion, $m$ is the mass of the sample, $c$ is the specific heat capacity of the sample, and $\Delta T$ is the change in temperature.

Calculating the Heat of Combustion

To calculate the heat of combustion, we need to know the mass of the sample, the specific heat capacity of the sample, and the change in temperature. The mass of the sample is given as 0.5 g. The specific heat capacity of $C_{10}H_8$ is 1.45 J/g°C.

The change in temperature can be calculated using the following equation:

$$\Delta T = T_f - T_i$$

where $T_f$ is the final temperature of the water and $T_i$ is the initial temperature of the water.

The Final Temperature of the Water

The final temperature of the water can be calculated using the following equation:

$$T_f = T_i + \frac{Q}{mc}$$

where $T_f$ is the final temperature of the water, $T_i$ is the initial temperature of the water, $Q$ is the heat of combustion, $m$ is the mass of the water, and $c$ is the specific heat capacity of the water.

Solving for the Final Temperature

To solve for the final temperature, we need to know the heat of combustion, the mass of the water, and the specific heat capacity of the water. The heat of combustion can be calculated using the equation:

$$Q = mc\Delta T$$

The mass of the water is given as 650 g. The specific heat capacity of water is 4.184 J/g°C.

Substituting Values

Substituting the values into the equation, we get:

$$Q = (0.5 \text{ g})(1.45 \text{ J/g°C})(\Delta T)$$

$$Q = 0.725 \text{ J/°C} \times \Delta T$$

The heat capacity of the calorimeter is given as $420 \frac{J}{{ }^{\circ} C}$. The heat capacity of the water is given as 4.184 J/g°C.

Solving for the Change in Temperature

To solve for the change in temperature, we need to know the heat of combustion, the heat capacity of the calorimeter, and the heat capacity of the water. The heat of combustion can be calculated using the equation:

$$Q = mc\Delta T$$

The heat capacity of the calorimeter is given as $420 \frac{J}{{ }^{\circ} C}$. The heat capacity of the water is given as 4.184 J/g°C.

Substituting Values

Substituting the values into the equation, we get:

$$Q = (0.5 \text{ g})(1.45 \text{ J/g°C})(\Delta T) + (420 \text{ J/°C})(\Delta T)$$

$$Q = 0.725 \text{ J/°C} \times \Delta T + 420 \text{ J/°C} \times \Delta T$$

$$Q = (0.725 + 420) \text{ J/°C} \times \Delta T$$

$$Q = 420.725 \text{ J/°C} \times \Delta T$$

Solving for the Final Temperature

To solve for the final temperature, we need to know the heat of combustion, the mass of the water, and the specific heat capacity of the water. The heat of combustion can be calculated using the equation:

$$Q = mc\Delta T$$

The mass of the water is given as 650 g. The specific heat capacity of water is 4.184 J/g°C.

Substituting Values

Substituting the values into the equation, we get:

$$Q = (650 \text{ g})(4.184 \text{ J/g°C})(\Delta T)$$

$$Q = 2718.4 \text{ J/°C} \times \Delta T$$

Equating the Two Equations

Equating the two equations, we get:

$$420.725 \text{ J/°C} \times \Delta T = 2718.4 \text{ J/°C} \times \Delta T$$

Solving for the Change in Temperature

To solve for the change in temperature, we need to divide both sides of the equation by the heat capacity of the calorimeter and the heat capacity of the water.

$$\Delta T = \frac{2718.4 \text{ J/°C}}{420.725 \text{ J/°C}}$$

$$\Delta T = 6.46^{\circ} C$$

Calculating the Final Temperature

The final temperature of the water can be calculated using the following equation:

$$T_f = T_i + \Delta T$$

where $T_f$ is the final temperature of the water, $T_i$ is the initial temperature of the water, and $\Delta T$ is the change in temperature.

Substituting Values

Substituting the values into the equation, we get:

$$T_f = 20.0^{\circ} C + 6.46^{\circ} C$$

$$T_f = 26.46^{\circ} C$$

Conclusion

In this article, we analyzed the burning of a 0.5 g sample of $C_{10}H_8$ in a calorimeter containing 650 g of water at $20.0^{\circ} C$. We calculated the heat of combustion of $C_{10}H_8$ and the final temperature of the water. The final temperature of the water was found to be $26.46^{\circ} C$.

References

• [1] CRC Handbook of Chemistry and Physics, 97th ed. (2016)
• [2] Kittel, C., Introduction to Solid State Physics, 8th ed. (2005)
• [3] Halliday, D., Resnick, R., Walker, J., Fundamentals of Physics, 9th ed. (2013)

Note

$$

Introduction

In our previous article, we analyzed the burning of a 0.5 g sample of $C_{10}H_8$ in a calorimeter containing 650 g of water at $20.0^{\circ} C$. We calculated the heat of combustion of $C_{10}H_8$ and the final temperature of the water. In this article, we will answer some of the most frequently asked questions related to this topic.

Q: What is the heat of combustion of $C_{10}H_8$?

A: The heat of combustion of $C_{10}H_8$ is a measure of the energy released during the combustion reaction. It can be calculated using the equation:

$$Q = mc\Delta T$$

where $Q$ is the heat of combustion, $m$ is the mass of the sample, $c$ is the specific heat capacity of the sample, and $\Delta T$ is the change in temperature.

Q: How do you calculate the final temperature of the water?

A: The final temperature of the water can be calculated using the following equation:

$$T_f = T_i + \Delta T$$

where $T_f$ is the final temperature of the water, $T_i$ is the initial temperature of the water, and $\Delta T$ is the change in temperature.

Q: What is the significance of the heat capacity of the calorimeter?

A: The heat capacity of the calorimeter is a measure of the amount of heat energy required to raise the temperature of the calorimeter by one degree Celsius. It is an important factor in calculating the heat of combustion of the sample.

Q: How do you calculate the heat capacity of the calorimeter?

A: The heat capacity of the calorimeter can be calculated using the following equation:

$$C = \frac{Q}{\Delta T}$$

where $C$ is the heat capacity of the calorimeter, $Q$ is the heat energy transferred to the calorimeter, and $\Delta T$ is the change in temperature.

Q: What is the difference between the heat of combustion and the heat capacity of the calorimeter?

A: The heat of combustion is a measure of the energy released during the combustion reaction, while the heat capacity of the calorimeter is a measure of the amount of heat energy required to raise the temperature of the calorimeter by one degree Celsius.

Q: How do you determine the mass of the sample?

A: The mass of the sample can be determined using a balance or a mass spectrometer.

Q: What is the significance of the specific heat capacity of the sample?

A: The specific heat capacity of the sample is a measure of the amount of heat energy required to raise the temperature of the sample by one degree Celsius.

Q: How do you calculate the specific heat capacity of the sample?

A: The specific heat capacity of the sample can be calculated using the following equation:

$$c = \frac{Q}{m\Delta T}$$

where $c$ is the specific heat capacity of the sample, $Q$ is the heat energy transferred to the sample, $m$ is the mass of the sample, and $\Delta T$ is the change in temperature.

Q: What is the difference between the specific heat capacity of the sample and the heat capacity of the calorimeter?

A: The specific heat capacity of the sample is a measure of the amount of heat energy required to raise the temperature of the sample by one degree Celsius, while the heat capacity of the calorimeter is a measure of the amount of heat energy required to raise the temperature of the calorimeter by one degree Celsius.

Conclusion

In this article, we answered some of the most frequently asked questions related to the burning of a 0.5 g sample of $C_{10}H_8$ in a calorimeter containing 650 g of water at $20.0^{\circ} C$. We hope that this article has provided a comprehensive understanding of the topic and has helped to clarify any doubts that you may have had.

References

• [1] CRC Handbook of Chemistry and Physics, 97th ed. (2016)
• [2] Kittel, C., Introduction to Solid State Physics, 8th ed. (2005)
• [3] Halliday, D., Resnick, R., Walker, J., Fundamentals of Physics, 9th ed. (2013)

Note

The values used in this article are approximate and may vary depending on the specific conditions of the experiment.