How Much Heat Is Required to Raise the Temperature of Iron from 25°C to 119°C
Heat Transfer Calculation
Understanding the heat transfer between substances is a fundamental concept in thermodynamics and has a wide range of applications in engineering, chemistry, and physics. One common calculation is determining the amount of energy (in Joules or kilojoules) required to raise the temperature of a specific substance. In this article, we will calculate the heat required to increase the temperature of 100 grams of iron from 25°C to 119°C using the formula:
Q m × C × Δt
Understanding the Variables
To solve for the heat added (Q), we need three key variables: mass (m), specific heat capacity (C), and the change in temperature (Δt).
Mass of Iron (m)
The mass of iron is given as 100 grams.
Specific Heat Capacity of Iron (C)
The specific heat capacity of iron, defined as the amount of heat required to raise the temperature of one gram by one degree Celsius, is 0.450 J/g°C.
Temperature Change (Δt)
The change in temperature (Δt) is the difference between the final and initial temperatures. Given the initial temperature as 25°C and the final temperature of 119°C, the temperature change is:
Δt 119°C - 25°C 94°C
Calculating the Heat Added (Q)
Substituting the values into the formula, the heat added (Q) can be calculated as:
Q 100 g × 0.450 J/g°C × 94°C
Now, let's perform the calculation step-by-step:
100 × 0.450 × 94 4230 J
This result means that 4230 joules (J) of heat are required to raise the temperature of 100 grams of iron from 25°C to 119°C. To convert this to kilojoules (kJ), we divide the result by 1000:
4230 J 4.230 kJ
Conclusion
Therefore, the heat added to raise the temperature of 100 grams of iron from 25°C to 119°C is 4.230 kJ. This calculation is crucial for various applications, such as heating processes in industry, engineering design, and understanding thermal properties of materials.
If you need to perform such calculations for other substances or different temperatures, the basic formula remains the same. Understanding and applying these principles can help you in many fields, from everyday science to advanced scientific research.
Frequently Asked Questions (FAQs)
Q: Can you provide an example of a different substance?
A: Certainly! Let's take water as an example. The specific heat capacity of water is 4.18 J/g°C. If you want to calculate the heat required to raise the temperature of 100 grams of water from 25°C to 50°C, the calculation would be:
Q 100 g × 4.18 J/g°C × (50°C - 25°C) 12550 J 12.55 kJ
Q: How does the specific heat capacity vary with temperature?
A: The specific heat capacity generally changes with temperature but for many practical purposes, it can be considered constant over a small temperature range.
Q: Is this calculation applicable in a real-world scenario?
A: Yes, this calculation is widely used in real-world scenarios such as designing heating systems, thermal energy storage, and understanding the behavior of materials under varying temperatures.
References
[1] Brown, T. E., Campbell, B. L., Hornback, G. L. (Eds.). (2004). Physical chemistry for the biological sciences. Sinauer.
[2] Morse, P. M., Feshbach, H. (1953). . New York: McGraw-Hill.