Energy transfer
The flow of energy from one region to another region is referred to as energy transfer. Since energy is quantitative; it must be transferred to a body or a material to work or to heat the system.
Molar Specific Heat
Heat capacity is the amount of heat energy absorbed or released by a chemical substance per the change in temperature of that substance. The change in heat is also called enthalpy. The SI unit of heat capacity is Joules per Kelvin, which is (J K-1)
Thermal Properties of Matter
Thermal energy is described as one of the form of heat energy which flows from one body of higher temperature to the other with the lower temperature when these two bodies are placed in contact to each other. Heat is described as the form of energy which is transferred between the two systems or in between the systems and their surrounding by the virtue of difference in temperature. Calorimetry is that branch of science which helps in measuring the changes which are taking place in the heat energy of a given body.
![**Problem: Calculating the Energy Required to Heat Copper**
The specific heat of copper is \( 385 \, \text{J/Kg/K} \). How much energy, in Joules, is needed to heat up a \( 0.4 \, \text{kg} \) of copper from \( 300 \, \text{Kelvin} \) to \( 450 \, \text{Kelvin} \)?
**Solution:**
To calculate the energy required to heat the copper, we use the formula for heat energy:
\[ Q = m \times c \times \Delta T \]
where:
- \( Q \) is the heat energy in Joules (J),
- \( m \) is the mass of the substance in kilograms (kg),
- \( c \) is the specific heat capacity in Joules per kilogram per Kelvin (\( \text{J/Kg/K} \)),
- \( \Delta T \) is the change in temperature in Kelvin (K).
Given:
- Specific heat capacity, \( c = 385 \, \text{J/Kg/K} \),
- Mass, \( m = 0.4 \, \text{kg} \),
- Initial temperature, \( T_i = 300 \, \text{K} \),
- Final temperature, \( T_f = 450 \, \text{K} \).
First, calculate the change in temperature:
\[ \Delta T = T_f - T_i = 450 \, \text{K} - 300 \, \text{K} = 150 \, \text{K} \]
Next, plug the values into the formula:
\[ Q = 0.4 \, \text{kg} \times 385 \, \text{J/Kg/K} \times 150 \, \text{K} \]
\[ Q = 0.4 \times 385 \times 150 \]
\[ Q = 23100 \, \text{J} \]
Therefore, the energy required to heat \( 0.4 \, \text{kg} \) of copper from \( 300 \, \text{K} \) to \( 450 \, \text{K} \) is \( 23100 \, \text{Joules} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbda06ec9-3160-4152-b2ee-882d8e600c29%2F534dbc8f-958e-41cb-8bbb-1cb35a4d948a%2Fj7bffnk_processed.png&w=3840&q=75)
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