Understanding Specific Heat
The specific heat capacity (often referred to simply as specific heat) is a physical property of materials. It varies from one substance to another and is instrumental in various practical applications, such as heating, cooling, and energy transfer processes.
Definition and Formula
Specific heat capacity can be defined mathematically using the formula:
\[
Q = mc\Delta T
\]
Where:
- \(Q\) = heat energy (in joules)
- \(m\) = mass of the substance (in kilograms)
- \(c\) = specific heat capacity (in joules per kilogram per degree Celsius, J/kg°C)
- \(\Delta T\) = change in temperature (in degrees Celsius)
This formula allows us to calculate the amount of heat energy absorbed or released by a substance when its temperature changes.
Common Values of Specific Heat
Different materials have unique specific heat values. Here are some common substances and their specific heat capacities:
- Water: 4.18 J/g°C
- Aluminum: 0.897 J/g°C
- Copper: 0.385 J/g°C
- Iron: 0.449 J/g°C
- Glass: 0.84 J/g°C
- Wood: 1.76 J/g°C
Practice Problems on Specific Heat
To solidify your understanding of specific heat, let's go through several practice problems. These problems will range in difficulty and will require you to apply the specific heat formula.
Example Problem 1: Heating Water
Problem Statement: Calculate the amount of heat required to raise the temperature of 200 grams of water from 20°C to 80°C.
Given:
- Mass of water (\(m\)) = 200 g = 0.2 kg (since 1 g = 0.001 kg)
- Specific heat of water (\(c\)) = 4.18 J/g°C = 4180 J/kg°C
- Initial temperature (\(T_1\)) = 20°C
- Final temperature (\(T_2\)) = 80°C
Solution:
1. Calculate \(\Delta T\):
\[
\Delta T = T_2 - T_1 = 80°C - 20°C = 60°C
\]
2. Apply the specific heat formula:
\[
Q = mc\Delta T = (0.2 \, \text{kg})(4180 \, \text{J/kg°C})(60°C)
\]
\[
Q = 0.2 \times 4180 \times 60 = 50280 \, \text{J}
\]
Answer: 50280 joules of heat is required.
Example Problem 2: Cooling Metal
Problem Statement: A copper block with a mass of 150 grams is initially at 100°C. If it cools down to 25°C, how much heat is released?
Given:
- Mass of copper (\(m\)) = 150 g = 0.15 kg
- Specific heat of copper (\(c\)) = 0.385 J/g°C = 385 J/kg°C
- Initial temperature (\(T_1\)) = 100°C
- Final temperature (\(T_2\)) = 25°C
Solution:
1. Calculate \(\Delta T\):
\[
\Delta T = T_2 - T_1 = 25°C - 100°C = -75°C
\]
2. Apply the specific heat formula:
\[
Q = mc\Delta T = (0.15 \, \text{kg})(385 \, \text{J/kg°C})(-75°C)
\]
\[
Q = 0.15 \times 385 \times -75 = -4331.25 \, \text{J}
\]
Answer: 4331.25 joules of heat is released.
Example Problem 3: Mixture of Substances
Problem Statement: If 500 grams of water at 25°C is mixed with 300 grams of aluminum at 80°C, what will be the final temperature of the mixture? Assume no heat is lost to the surroundings.
Given:
- Mass of water (\(m_w\)) = 500 g = 0.5 kg
- Specific heat of water (\(c_w\)) = 4180 J/kg°C
- Initial temperature of water (\(T_{w1}\)) = 25°C
- Mass of aluminum (\(m_{Al}\)) = 300 g = 0.3 kg
- Specific heat of aluminum (\(c_{Al}\)) = 897 J/kg°C
- Initial temperature of aluminum (\(T_{Al1}\)) = 80°C
Solution:
Let \(T_f\) be the final temperature of the mixture. The heat lost by aluminum will equal the heat gained by water.
\[
m_{Al}c_{Al}(T_{Al1} - T_f) = m_w c_w (T_f - T_{w1})
\]
Substituting the values:
\[
(0.3)(897)(80 - T_f) = (0.5)(4180)(T_f - 25)
\]
Expanding both sides:
\[
269.1(80 - T_f) = 2090(T_f - 25)
\]
Distributing:
\[
21528 - 269.1T_f = 2090T_f - 52250
\]
Combining like terms:
\[
21528 + 52250 = 2090T_f + 269.1T_f
\]
\[
73778 = 2359.1T_f
\]
Solving for \(T_f\):
\[
T_f = \frac{73778}{2359.1} \approx 31.25°C
\]
Answer: The final temperature of the mixture is approximately 31.25°C.
Conclusion
Understanding specific heat and practicing related problems are crucial for mastering concepts in thermodynamics. By grasping how to calculate heat transfer in various scenarios, students can apply this knowledge to real-world situations, from cooking to industrial processes. Working through specific heat practice problems not only reinforces theoretical concepts but also builds problem-solving skills essential in scientific disciplines. Remember to always consider the units you're working with and ensure that they are consistent throughout your calculations.
Frequently Asked Questions
What is specific heat and why is it important in thermodynamics?
Specific heat is the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius. It is important in thermodynamics because it helps predict how substances will react to heat energy, which is crucial in various applications including heating, cooling, and material selection.
How do you calculate the heat absorbed by a substance using specific heat?
The heat absorbed can be calculated using the formula Q = mcΔT, where Q is the heat absorbed, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
If 200 grams of water is heated from 25°C to 75°C, how much heat is absorbed?
Using the specific heat of water (4.18 J/g°C), the heat absorbed can be calculated as Q = mcΔT = 200 g 4.18 J/g°C (75°C - 25°C) = 41800 J or 41.8 kJ.
What factors affect the specific heat of a substance?
Factors that affect specific heat include the phase of the substance (solid, liquid, gas), temperature, and the molecular structure or composition of the material.
How can you determine the specific heat of an unknown metal using a calorimeter?
To determine the specific heat of an unknown metal, you can heat the metal to a known temperature, place it in a calorimeter with water at a known temperature, and measure the final equilibrium temperature. Using the principle of conservation of energy, you can solve for the specific heat of the metal.
What is the specific heat of aluminum, and how can it be used in practice problems?
The specific heat of aluminum is approximately 0.897 J/g°C. In practice problems, it can be used to calculate the heat transfer when aluminum is heated or cooled, such as in manufacturing processes or thermal management applications.
How does specific heat affect the heating time of different materials?
Materials with a low specific heat will heat up and cool down more quickly than those with a high specific heat. This is important in applications like cooking, where different materials require different heating times to achieve the desired temperature.
Can specific heat be different for the same substance at different temperatures?
Yes, specific heat can vary with temperature due to changes in the physical state or molecular interactions of the substance. In practice problems, it is important to use the specific heat value that corresponds to the temperature range of interest.
