t is the specific heat of a metal substance if 135 kJ of heat is needed to raise metal from 18.0°C to 37.2°C?

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### Physics Problems on Thermal Energy and Specific Heat

#### Problem 5:
**(14.10)** What is the specific heat of a metal substance if 135 kJ of heat is needed to raise 4.1 kg of the metal from 18.0°C to 37.2°C?

---

#### Problem 6:
**(14.21)** A 0.095-kg aluminum sphere is dropped from the roof of a 55-m-high building. If 65% of the thermal energy produced when it hits the ground is absorbed by the sphere, what is its temperature increase?

---

### Explanation of Concepts:

**Specific Heat (Problem 5):**
The specific heat capacity of a substance is a measure of how much heat energy (in joules or kilojoules) is required to raise the temperature of a unit mass (typically 1 kg) of the substance by 1 degree Celsius (°C). 

The formula for specific heat is given by:

\[ 
Q = mc\Delta T 
\]

where:
- \( Q \) is the heat energy added (in kilojoules or joules)
- \( m \) is the mass of the substance (in kilograms)
- \( c \) is the specific heat capacity (in kJ/kg°C or J/kg°C)
- \( \Delta T \) is the change in temperature (in °C)

**Thermal Energy and Temperature Increase (Problem 6):**
When an object falls from a height and impacts the ground, the potential energy it loses is converted into other forms of energy, such as thermal energy. The change in temperature due to the absorbed energy can be calculated using specific heat principles.

The potential energy lost by an object of mass \( m \) falling from height \( h \) is given by:

\[ 
PE = mgh 
\]

where:
- \( g \) is the acceleration due to gravity (approximately 9.8 m/s²)

If a percentage of this energy is converted into thermal energy absorbed by the object, this energy can be used to determine the temperature change using the specific heat formula mentioned above.

---

These problems help in understanding the concepts of specific heat and the transformation of potential energy into thermal energy.
Transcribed Image Text:Certainly! Here is a transcription and explanation suitable for an educational website: --- ### Physics Problems on Thermal Energy and Specific Heat #### Problem 5: **(14.10)** What is the specific heat of a metal substance if 135 kJ of heat is needed to raise 4.1 kg of the metal from 18.0°C to 37.2°C? --- #### Problem 6: **(14.21)** A 0.095-kg aluminum sphere is dropped from the roof of a 55-m-high building. If 65% of the thermal energy produced when it hits the ground is absorbed by the sphere, what is its temperature increase? --- ### Explanation of Concepts: **Specific Heat (Problem 5):** The specific heat capacity of a substance is a measure of how much heat energy (in joules or kilojoules) is required to raise the temperature of a unit mass (typically 1 kg) of the substance by 1 degree Celsius (°C). The formula for specific heat is given by: \[ Q = mc\Delta T \] where: - \( Q \) is the heat energy added (in kilojoules or joules) - \( m \) is the mass of the substance (in kilograms) - \( c \) is the specific heat capacity (in kJ/kg°C or J/kg°C) - \( \Delta T \) is the change in temperature (in °C) **Thermal Energy and Temperature Increase (Problem 6):** When an object falls from a height and impacts the ground, the potential energy it loses is converted into other forms of energy, such as thermal energy. The change in temperature due to the absorbed energy can be calculated using specific heat principles. The potential energy lost by an object of mass \( m \) falling from height \( h \) is given by: \[ PE = mgh \] where: - \( g \) is the acceleration due to gravity (approximately 9.8 m/s²) If a percentage of this energy is converted into thermal energy absorbed by the object, this energy can be used to determine the temperature change using the specific heat formula mentioned above. --- These problems help in understanding the concepts of specific heat and the transformation of potential energy into thermal energy.
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