A 4.50 G Coin Of Copper Absorbed 45 Calories Of Heat. What Was The Final Temperature Of The Copper If The Initial Temperature Was $25^{\circ} C$? The Specific Heat Of Copper Is $0.092 \, \text{cal/g}^{\circ} C$.

Introduction

In this problem, we are given a 4.50 g coin of copper that absorbs 45 calories of heat. We need to find the final temperature of the copper, given that the initial temperature is $25^{\circ} C$ and the specific heat of copper is $0.092 \, \text{cal/g}^{\circ} C$. This problem involves the concept of specific heat capacity, which is a measure of the amount of heat energy required to raise the temperature of a substance by one degree Celsius.

Understanding Specific Heat Capacity

Specific heat capacity is defined as the amount of heat energy required to raise the temperature of a substance by one degree Celsius. It is typically denoted by the symbol $c$ and has units of $\text{cal/g}^{\circ} C$. The specific heat capacity of a substance depends on its chemical composition and physical properties.

Calculating the Final Temperature

To calculate the final temperature of the copper, we can use the formula:

$$Q = mc\Delta T$$

where $Q$ is the amount of heat energy absorbed, $m$ is the mass of the substance, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature.

Rearranging the formula to solve for $\Delta T$, we get:

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

Substituting the given values, we get:

$$\Delta T = \frac{45 \, \text{cal}}{(4.50 \, \text{g})(0.092 \, \text{cal/g}^{\circ} C)}$$

Simplifying the expression, we get:

$$\Delta T = 100^{\circ} C$$

Since the initial temperature is $25^{\circ} C$, the final temperature is:

$$T_f = T_i + \Delta T = 25^{\circ} C + 100^{\circ} C = 125^{\circ} C$$

Conclusion

In this problem, we calculated the final temperature of a 4.50 g coin of copper that absorbed 45 calories of heat. We used the formula $Q = mc\Delta T$ to solve for the change in temperature, and then added the initial temperature to find the final temperature. The final temperature of the copper is $125^{\circ} C$.

Key Concepts

• Specific heat capacity: the amount of heat energy required to raise the temperature of a substance by one degree Celsius.
• Formula: $Q = mc\Delta T$.
• Rearranging the formula to solve for $\Delta T$: $\Delta T = \frac{Q}{mc}$.
• Calculating the final temperature: $T_f = T_i + \Delta T$.

Real-World Applications

This problem has real-world applications in various fields, such as:

• Thermodynamics: understanding the behavior of heat energy and its effects on substances.
• Materials Science: studying the properties of materials and their responses to heat energy.
• Engineering: designing systems that involve heat transfer and temperature changes.

Additional Resources

For further learning, you can explore the following resources:

• Textbooks: "Thermodynamics" by C. J. Adkins, "Materials Science" by William D. Callister Jr.
• Online Courses: "Thermodynamics" on Coursera, "Materials Science" on edX.
• Websites: American Society of Mechanical Engineers (ASME), Materials Science and Engineering (MSE) community.
  A 4.50 g Coin of Copper Absorbs Heat: Calculating the Final Temperature ===========================================================

Q&A: Frequently Asked Questions

Q: What is specific heat capacity?

A: Specific heat capacity is the amount of heat energy required to raise the temperature of a substance by one degree Celsius. It is typically denoted by the symbol $c$ and has units of $\text{cal/g}^{\circ} C$.

Q: How is specific heat capacity related to the formula $Q = mc\Delta T$?

A: The formula $Q = mc\Delta T$ shows that specific heat capacity ($c$) is a measure of the amount of heat energy required to raise the temperature of a substance by one degree Celsius. The formula can be rearranged to solve for $\Delta T$, which is the change in temperature.

Q: What is the significance of the initial temperature in this problem?

A: The initial temperature is the starting temperature of the copper coin before it absorbs heat energy. It is used as a reference point to calculate the final temperature of the copper coin.

Q: How is the final temperature calculated?

A: The final temperature is calculated by adding the change in temperature ($\Delta T$) to the initial temperature ($T_i$). In this problem, the final temperature is $T_f = T_i + \Delta T = 25^{\circ} C + 100^{\circ} C = 125^{\circ} C$.

Q: What are some real-world applications of this problem?

A: This problem has real-world applications in various fields, such as:

• Thermodynamics: understanding the behavior of heat energy and its effects on substances.
• Materials Science: studying the properties of materials and their responses to heat energy.
• Engineering: designing systems that involve heat transfer and temperature changes.

Q: What are some additional resources for further learning?

A: For further learning, you can explore the following resources:

• Textbooks: "Thermodynamics" by C. J. Adkins, "Materials Science" by William D. Callister Jr.
• Online Courses: "Thermodynamics" on Coursera, "Materials Science" on edX.
• Websites: American Society of Mechanical Engineers (ASME), Materials Science and Engineering (MSE) community.

Q: What is the significance of the specific heat capacity of copper in this problem?

A: The specific heat capacity of copper ($0.092 \, \text{cal/g}^{\circ} C$) is a measure of the amount of heat energy required to raise the temperature of copper by one degree Celsius. It is used in the formula $Q = mc\Delta T$ to calculate the change in temperature.

Q: How does the mass of the copper coin affect the calculation of the final temperature?

A: The mass of the copper coin affects the calculation of the final temperature by changing the value of the term $mc$ in the formula $Q = mc\Delta T$. A larger mass of copper would result in a larger value of $mc$, which would lead to a larger change in temperature.

Q: What are some common mistakes to avoid when solving this problem?

A: Some common mistakes to avoid when solving this problem include:

• Incorrectly calculating the change in temperature: make sure to use the correct formula and units.
• Forgetting to add the initial temperature: make sure to add the initial temperature to the change in temperature to get the final temperature.
• Using the wrong value for the specific heat capacity: make sure to use the correct value for the specific heat capacity of copper.