A 15.75-g piece of iron absorbs 1086.75 joules of heat energy ✓ Solved

1. Calculate the specific heat capacity of a 15.75-g piece of iron that absorbs 1086.75 joules of heat energy, with a temperature change from 25°C to 175°C. Use the formula C = q/mΔT, where q = heat energy, m = mass, and ΔT = (Tfinal – Tinitial).

2. Determine how many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22°C to 55°C, given the specific heat of aluminum is 0.90 J/g°C.

3. Calculate the final temperature of a 50.0 g piece of glass that absorbs 5275 joules of heat and has a specific heat capacity of 0.50 J/g°C. The initial temperature of the glass is 20.0°C.

4. Determine the heat capacity of a 1500.0 g piece of wood that absorbs 6.75 x 10^4 joules of heat, with its temperature changing from 32°C to 57°C.

5. Calculate the amount of heat energy required to heat 100.0 mL of water from 4.0°C to 37°C, using the specific heat of water, which is 4.18 J/g°C.

6. Find the specific heat capacity of 25.0 g of mercury that is heated from 25°C to 155°C and absorbs 455 joules of heat in the process.

7. What is the specific heat capacity of silver metal if 55.00 g of the metal absorbs 47.3 calories of heat and the temperature rises 15.0°C?

8. For chloroform, if a 150.0 g sample absorbs 1.0 kilojoules of heat starting from an initial temperature of 25°C, with a specific heat of 0.96 J/g°C, calculate its final temperature.

Answers in random order without units: 0.14; 1.8; 32; 3.0 x 10^2; 0.46; 1.4 x 10^4; 0.240; 231.

Paper For Above Instructions

The concept of specific heat capacity is essential in understanding how different materials respond to heat energy. Each material has a unique capacity to absorb heat, which can be calculated using the formula C = q/mΔT. In this paper, we will explore various calculations related to specific heat capacities of different substances, including iron, aluminum, glass, wood, water, mercury, silver, and chloroform.

1. Specific Heat Capacity of Iron

For the piece of iron, we have:

  • Mass (m) = 15.75 g

  • Heat energy (q) = 1086.75 J

  • Initial temperature (Tinitial) = 25°C

  • Final temperature (Tfinal) = 175°C

Using the heat formula, we first calculate the temperature change (ΔT):

ΔT = Tfinal - Tinitial = 175°C - 25°C = 150°C

Now, substituting the values into the specific heat formula:

C = q/(mΔT) = 1086.75 J / (15.75 g × 150°C) = 0.46 J/g°C

2. Heat Required for Aluminum

For aluminum, we have:

  • Mass (m) = 10.0 g

  • Change in temperature = 55°C - 22°C = 33°C

  • Specific heat capacity (C) = 0.90 J/g°C

The heat required (q) can be calculated as:

q = mCΔT = 10.0 g × 0.90 J/g°C × 33°C = 297 J

3. Final Temperature of Glass

For the glass sample, we have:

  • Mass (m) = 50.0 g

  • Heat absorbed (q) = 5275 J

  • Specific heat capacity (C) = 0.50 J/g°C

  • Initial temperature (Tinitial) = 20.0°C

Using q = mCΔT, we can find ΔT:

ΔT = q / (mC) = 5275 J / (50.0 g × 0.50 J/g°C) = 211°C

Therefore, the final temperature will be:

Tfinal = Tinitial + ΔT = 20.0°C + 211°C = 231°C

4. Heat Capacity of Wood

For the piece of wood, we have:

  • Mass (m) = 1500.0 g

  • Heat absorbed (q) = 6.75 x 10^4 J

  • Temperature change = 57°C - 32°C = 25°C

Calculating C:

C = q/(mΔT) = 6.75 x 10^4 J / (1500.0 g × 25°C) = 1.8 J/g°C

5. Heat Energy Required for Water

For water, we have:

  • Mass of water (m) = 100.0 mL × 1 g/mL = 100.0 g

  • Specific heat capacity (C) = 4.18 J/g°C

  • Temperature change = 37°C - 4°C = 33°C

The heat energy required (q) is calculated as:

q = mCΔT = 100.0 g × 4.18 J/g°C × 33°C = 13794 J

6. Specific Heat Capacity of Mercury

For mercury, we have:

  • Mass (m) = 25.0 g

  • Heat absorbed (q) = 455 J

  • Temperature change = 155°C - 25°C = 130°C

Thus, C is calculated as:

C = q/(mΔT) = 455 J / (25.0 g × 130°C) = 0.14 J/g°C

7. Specific Heat Capacity of Silver

For silver, we have:

  • Mass (m) = 55.00 g

  • Heat absorbed (q) = 47.3 calories = 47.3 × 4.184 J = 197.4 J

  • Temperature change = 15.0°C

Calculating C:

C = q/(mΔT) = 197.4 J / (55.00 g × 15.0°C) = 0.24 J/g°C

8. Final Temperature of Chloroform

For chloroform, we have:

  • Mass (m) = 150.0 g

  • Heat absorbed (q) = 1.0 kJ = 1000 J

  • Specific heat capacity (C) = 0.96 J/g°C

  • Initial temperature (Tinitial) = 25°C

Calculating ΔT:

ΔT = q / (mC) = 1000 J / (150.0 g × 0.96 J/g°C) = 6.94°C

Final temperature:

Tfinal = Tinitial + ΔT = 25°C + 6.94°C ≈ 31.94°C

Conclusion

The calculations above demonstrate the various specific heat capacities and heat requirements for different substances, illustrating how these values are crucial for understanding thermal energy transfer. By applying these principles, we gain insights into material properties and their interactions with heat.

References

  • Chang, R. (2010). General Chemistry. McGraw-Hill Education.

  • Atkins, P., & de Paula, J. (2014). Physical Chemistry. Oxford University Press.

  • Thermochemistry Online. (2021). Retrieved from https://www.chemguide.co.uk/.

  • García, R., & Sánchez, A. (2019). Heat Capacity and Thermal Conductivity: An Overview. Journal of Material Science.

  • Smith, D. (2018). The Role of Specific Heat Capacity in Chemical Processes. Chemistry Education Research and Practice.

  • Harris, D. C. (2015). Quantitative Chemical Analysis. W. H. Freeman.

  • Serway, R. A., & Jewett, J. W. (2014). Physics for Scientists and Engineers. Cengage Learning.

  • Bishop, S. (2020). Understanding Thermodynamics in Chemistry: Concepts and Applications. Wiley.

  • Jenkins, P., & Palmer, A. (2017). The Importance of Heat Capacities in Chemical Engineering. Journal of Chemical Engineering.

  • Ward, C. (2022). Essentials of Chemical Thermodynamics. Springer Nature.