Specific Heat Capacity Calculations Explained

Introduction to Specific Heat Capacity Calculations

Specific heat capacity is a crucial concept in thermochemistry that describes the amount of heat required to change the temperature of a substance. It is typically expressed in joules per gram per degree Celsius (J/g°C). Understanding how to calculate specific heat capacity is vital in various applications, such as heating and cooling processes, material selection, and energy efficiency assessments. In this article, we will explore three diverse examples of specific heat capacity calculations to illustrate its practical applications.

Example 1: Heating Water in a Kettle

In a typical household scenario, when boiling water in a kettle, it is essential to understand how much energy is needed to raise its temperature. Let’s say you want to heat 500 grams of water from room temperature (25°C) to boiling point (100°C). Water has a specific heat capacity of approximately 4.18 J/g°C.
To find the amount of energy required, use the formula:

\[ Q = m imes c imes \Delta T \]
Where:

• \( Q \) = heat energy (in joules)
• \( m \) = mass (in grams)
• \( c \) = specific heat capacity (in J/g°C)
• \( \Delta T \) = change in temperature (in °C)

In this case:

• Mass \( m = 500 \, \text{g} \)
• Specific heat capacity \( c = 4.18 \, \text{J/g°C} \)
• Change in temperature \( \Delta T = 100 - 25 = 75 \, °C \)

Now, substituting the values into the formula:

\[ Q = 500 \, ext{g} imes 4.18 \, ext{J/g°C} imes 75 \, °C = 156750 \, ext{J} \]

Thus, it requires 156,750 joules of energy to heat the water from 25°C to 100°C.

Notes:

• This calculation assumes no heat loss to the environment, which is idealized. In reality, some energy will be lost during heating.

Example 2: Cooling a Metal Cube in Ice Water

Consider a scenario where you have a metal cube weighing 300 grams at a temperature of 150°C, and you want to cool it down by immersing it in ice water (0°C). Let’s assume the specific heat capacity of the metal is 0.385 J/g°C.
To determine how much heat is lost by the metal cube as it cools down, we can again use the formula:

\[ Q = m imes c imes \Delta T \]

Here:

• Mass \( m = 300 \, \text{g} \)
• Specific heat capacity \( c = 0.385 \, ext{J/g°C} \)
• Change in temperature \( \Delta T = 0 - 150 = -150 \, °C \)

Substituting the values:

\[ Q = 300 \, ext{g} imes 0.385 \, ext{J/g°C} imes -150 \, °C = -17325 \, ext{J} \]

The negative sign indicates that the metal cube is losing heat. Therefore, it releases 17,325 joules of energy while cooling down.

Notes:

• The final temperature of the metal will depend on the heat exchange with the ice water and its specific heat capacity.

Example 3: Heating a Block of Ice

Let’s assess a situation where you have a 200-gram block of ice at -10°C, and you want to heat it until it melts and reaches 0°C. The specific heat capacity of ice is about 2.09 J/g°C. Using the specific heat formula:

\[ Q = m imes c imes \Delta T \]

In this scenario:

• Mass \( m = 200 \, ext{g} \)
• Specific heat capacity \( c = 2.09 \, ext{J/g°C} \)
• Change in temperature \( \Delta T = 0 - (-10) = 10 \, °C \)

Substituting these values:

\[ Q = 200 \, ext{g} imes 2.09 \, ext{J/g°C} imes 10 \, °C = 4180 \, ext{J} \]

Thus, it requires 4,180 joules of energy to heat the ice from -10°C to 0°C before it starts to melt.

Notes:

• To calculate the total energy needed to fully melt the ice afterward, you would also need to consider the latent heat of fusion, which is an additional calculation.

These examples illustrate the application of specific heat capacity calculations in real-world scenarios, providing an essential understanding of thermal energy transfer.