When 25.0 grams of water are cooled from 20.0 degrees Celsius to 10.0 degrees Celsius the number of joules of heat energy released is?
The Correct Answer and Explanation is :
To calculate the amount of heat energy released when cooling a given mass of water, we can use the formula:
[
q = m \cdot c \cdot \Delta T
]
Where:
- ( q ) is the heat energy (in joules)
- ( m ) is the mass of the substance (in grams)
- ( c ) is the specific heat capacity of the substance (for water, it’s approximately ( 4.18 \, \text{J/g°C} ))
- ( \Delta T ) is the change in temperature (in °C)
Step-by-Step Calculation
- Identify the Variables:
- Mass of water (( m )) = 25.0 grams
- Specific heat capacity of water (( c )) = 4.18 J/g°C
- Initial temperature (( T_i )) = 20.0 °C
- Final temperature (( T_f )) = 10.0 °C
- Calculate the Temperature Change (( \Delta T )):
[
\Delta T = T_f – T_i = 10.0 \, \text{°C} – 20.0 \, \text{°C} = -10.0 \, \text{°C}
]
The negative sign indicates that the temperature is decreasing. - Plug the Values into the Formula:
[
q = m \cdot c \cdot \Delta T
]
[
q = 25.0 \, \text{g} \cdot 4.18 \, \text{J/g°C} \cdot (-10.0 \, \text{°C})
]
[
q = 25.0 \cdot 4.18 \cdot (-10.0) = -1045 \, \text{J}
]
Conclusion
The heat energy released when 25.0 grams of water is cooled from 20.0 degrees Celsius to 10.0 degrees Celsius is approximately 1045 joules. The negative sign indicates that heat is being released to the surroundings as the water cools down. This process illustrates the principle of conservation of energy, where energy is transferred out of the water as it loses thermal energy, which is fundamental in thermodynamics.