Answer :
To determine the amount of heat released when an aluminum skillet cools down, we can use the formula for heat transfer:
[tex]\[ Q = mc\Delta T \][/tex]
Where:
- [tex]\( Q \)[/tex] is the heat released (in joules).
- [tex]\( m \)[/tex] is the mass of the skillet (in grams).
- [tex]\( c \)[/tex] is the specific heat capacity of the material (in joules per gram degree Celsius).
- [tex]\( \Delta T \)[/tex] is the change in temperature (in degrees Celsius).
Given data:
- Mass of the skillet, [tex]\( m = 1580 \)[/tex] g
- Initial temperature, [tex]\( T_{\text{initial}} = 173^{\circ} C \)[/tex]
- Final temperature, [tex]\( T_{\text{final}} = 23.9^{\circ} C \)[/tex]
- Specific heat capacity of aluminum, [tex]\( c = 0.901 \, \text{J/g} \cdot ^{\circ} C \)[/tex]
First, we need to find the change in temperature [tex]\( \Delta T \)[/tex]:
[tex]\[ \Delta T = T_{\text{initial}} - T_{\text{final}} \][/tex]
[tex]\[ \Delta T = 173^{\circ} C - 23.9^{\circ} C \][/tex]
[tex]\[ \Delta T = 149.1^{\circ} C \][/tex]
Next, substitute the values into the heat transfer formula:
[tex]\[ Q = mc\Delta T \][/tex]
[tex]\[ Q = 1580 \, \text{g} \times 0.901 \, \text{J/g} \cdot ^{\circ} C \times 149.1^{\circ} C \][/tex]
So,
[tex]\[ Q = 1580 \times 0.901 \times 149.1 \][/tex]
[tex]\[ Q = 212255.778 \, \text{J} \][/tex]
Therefore, the amount of heat released when the aluminum skillet cools down to room temperature is [tex]\( 212255.778 \)[/tex] joules.
[tex]\[ Q = mc\Delta T \][/tex]
Where:
- [tex]\( Q \)[/tex] is the heat released (in joules).
- [tex]\( m \)[/tex] is the mass of the skillet (in grams).
- [tex]\( c \)[/tex] is the specific heat capacity of the material (in joules per gram degree Celsius).
- [tex]\( \Delta T \)[/tex] is the change in temperature (in degrees Celsius).
Given data:
- Mass of the skillet, [tex]\( m = 1580 \)[/tex] g
- Initial temperature, [tex]\( T_{\text{initial}} = 173^{\circ} C \)[/tex]
- Final temperature, [tex]\( T_{\text{final}} = 23.9^{\circ} C \)[/tex]
- Specific heat capacity of aluminum, [tex]\( c = 0.901 \, \text{J/g} \cdot ^{\circ} C \)[/tex]
First, we need to find the change in temperature [tex]\( \Delta T \)[/tex]:
[tex]\[ \Delta T = T_{\text{initial}} - T_{\text{final}} \][/tex]
[tex]\[ \Delta T = 173^{\circ} C - 23.9^{\circ} C \][/tex]
[tex]\[ \Delta T = 149.1^{\circ} C \][/tex]
Next, substitute the values into the heat transfer formula:
[tex]\[ Q = mc\Delta T \][/tex]
[tex]\[ Q = 1580 \, \text{g} \times 0.901 \, \text{J/g} \cdot ^{\circ} C \times 149.1^{\circ} C \][/tex]
So,
[tex]\[ Q = 1580 \times 0.901 \times 149.1 \][/tex]
[tex]\[ Q = 212255.778 \, \text{J} \][/tex]
Therefore, the amount of heat released when the aluminum skillet cools down to room temperature is [tex]\( 212255.778 \)[/tex] joules.
