Understanding specific heat capacity is crucial for a variety of scientific and engineering applications. This concept, often encountered in thermodynamics and physics, describes the amount of heat required to raise the temperature of a substance. Let's dive deep into how to solve common specific heat problems by looking at a worksheet that focuses on this topic.
Understanding Specific Heat Capacity
Specific heat capacity (c) is defined as the amount of heat energy required to raise the temperature of one unit mass of a substance by one degree Celsius or Kelvin. The formula used to calculate specific heat capacity is:
[ q = mc\Delta T ]
- q: heat energy (J)
- m: mass of the substance (kg)
- c: specific heat capacity (J/kg°C or J/kg·K)
- \Delta T: change in temperature (°C or K)
Example 1: Calculating Heat Energy
A sample of aluminum with a mass of 0.5 kg is heated from 20°C to 100°C. Given that the specific heat capacity of aluminum is 900 J/kg·°C, calculate the heat energy absorbed.
- Mass, m = 0.5 kg
- Specific heat capacity, c = 900 J/kg·°C
- Change in temperature, \Delta T = 100°C - 20°C = 80°C
Using the formula:
[ q = 0.5 \, kg \times 900 \, \frac{J}{kg \, \cdot °C} \times 80 \, °C = 36,000 \, J ]
Thus, the heat energy absorbed is 36,000 Joules.
🔬 Note: Always ensure units are consistent when calculating. If temperature change is in Celsius, make sure specific heat capacity matches in units.
Example 2: Finding Specific Heat Capacity
300 grams of a substance is heated, absorbing 2400 Joules of heat, and the temperature rises by 12°C. What is the specific heat capacity of the substance?
- Mass, m = 300 g = 0.3 kg
- Heat energy, q = 2400 J
- Change in temperature, \Delta T = 12°C
Solve for c:
[ c = \frac{q}{m \Delta T} = \frac{2400 \, J}{0.3 \, kg \times 12 \, °C} = 666.67 \, J/kg·°C ]
The specific heat capacity of the substance is approximately 667 J/kg·°C.
💡 Note: The number of significant figures should be consistent with the least precise data given.
Example 3: Temperature Change
100 grams of water at 25°C absorbs 41,840 Joules of heat. What is its final temperature if the specific heat of water is 4.184 J/g·°C?
- Mass, m = 100 g = 0.1 kg
- Heat energy, q = 41,840 J
- Specific heat capacity, c = 4.184 J/g·°C
Solve for \Delta T:
[ \Delta T = \frac{q}{mc} = \frac{41,840 \, J}{0.1 \, kg \times 4.184 \, \frac{J}{g \, \cdot °C} \times 1000 \frac{g}{kg}} = 100 \, °C ]
The final temperature of the water will be 125°C.
🚫 Note: If the specific heat units don't match, convert accordingly for consistency.
Example 4: Mixing Two Substances
A 0.2 kg piece of copper at 100°C is placed into 1.0 kg of water at 20°C. What will be the final temperature when thermal equilibrium is reached? (Specific heat capacity of copper = 385 J/kg·°C)
| Substance | Mass (kg) | Initial Temp. (°C) | Specific Heat (J/kg·°C) |
|---|---|---|---|
| Copper | 0.2 | 100 | 385 |
| Water | 1.0 | 20 | 4184 |
We use the principle of heat conservation:
[ mc{\text{copper}} \Delta T{\text{copper}} = mc{\text{water}} \Delta T{\text{water}} ]
[ 0.2 \times 385 \times (100 - T_f) = 1.0 \times 4184 \times (T_f - 20) ]
Solving this equation yields:
[ T_f \approx 23.1°C ]
The final temperature when equilibrium is reached will be approximately 23.1°C.
Example 5: Phase Change
100 grams of ice at -10°C is heated to become steam at 100°C. What is the heat required if specific heat capacity of ice is 2.09 J/g·°C, water 4.184 J/g·°C, and the latent heats for melting and vaporization are 334 J/g and 2260 J/g respectively?
- Heat to warm ice from -10°C to 0°C: q_1 = 100 \times 2.09 \times 10 = 2090\, J
- Heat to melt ice to water at 0°C: q_2 = 100 \times 334 = 33400\, J
- Heat to warm water from 0°C to 100°C: q_3 = 100 \times 4.184 \times 100 = 418400\, J
- Heat to turn water into steam: q_4 = 100 \times 2260 = 226000\, J
Total heat required:
[ q_{\text{total}} = q_1 + q_2 + q_3 + q_4 \approx 678,490 \, J ]
💦 Note: Phase changes involve latent heat, not specific heat, and should be accounted for separately.
In this detailed exploration of specific heat capacity through various examples, we've seen how this physical property plays a fundamental role in heat transfer. From simple calculations to more complex scenarios involving phase changes, understanding specific heat enables precise predictions in energy requirements, efficiency, and material behavior under thermal conditions. This knowledge not only helps in solving problems but also in designing systems where heat management is critical.
What is the difference between specific heat and heat capacity?
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Specific heat capacity is the amount of heat needed to raise the temperature of one unit mass of a substance by one degree, while heat capacity refers to the total heat required to raise the temperature of the entire object by one degree.
How does the specific heat of a substance affect its use in thermal systems?
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A substance with high specific heat can absorb or release more heat without a significant change in temperature, making it ideal for thermal storage or insulation. Conversely, materials with low specific heat are better for quick heating or cooling applications.
Why does water have such a high specific heat capacity?
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Water’s high specific heat capacity comes from hydrogen bonding. These bonds absorb heat energy to break, allowing more energy to be stored as heat without a large increase in temperature.
How do changes in phase (like melting or vaporization) affect heat calculations?
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Phase changes involve latent heat which is the heat required to change the state of a substance without changing its temperature. This heat must be added to or removed from the system separately from the heat that changes the temperature of the substance.
Can specific heat capacity change with temperature?
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Yes, specific heat capacity can vary with temperature, especially at extreme temperatures. However, within moderate ranges and for most common applications, it’s often considered constant.
