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Sagot :
To determine the energy lost by the water when a cold piece of iron is placed in it, we can follow these steps:
1. Identify the given data:
- Mass of water ([tex]\( m \)[/tex]): 100 grams
- Initial temperature of water ([tex]\( T_i \)[/tex]): 23.4°C
- Final temperature of water ([tex]\( T_f \)[/tex]): 20.9°C
- Specific heat capacity of water ([tex]\( c \)[/tex]): 4.18 J/g°C
2. Calculate the change in temperature ([tex]\( \Delta T \)[/tex]):
[tex]\[ \Delta T = T_i - T_f \][/tex]
Substituting the values:
[tex]\[ \Delta T = 23.4°C - 20.9°C = 2.5°C \][/tex]
3. Apply the formula to calculate the energy lost ([tex]\( Q \)[/tex]):
The formula to calculate the energy lost or gained is:
[tex]\[ Q = mc\Delta T \][/tex]
Where:
- [tex]\( Q \)[/tex] is the energy transferred,
- [tex]\( m \)[/tex] is the mass of the water,
- [tex]\( c \)[/tex] is the specific heat capacity,
- [tex]\( \Delta T \)[/tex] is the change in temperature.
4. Substitute the known values into the formula:
[tex]\[ Q = 100 \, \text{g} \times 4.18 \, \text{J/g°C} \times 2.5 \, \text{°C} \][/tex]
5. Perform the multiplication:
[tex]\[ Q = 100 \times 4.18 \times 2.5 = 1045.0 \, \text{J} \][/tex]
So, the energy lost by the water to the iron is 1045.0 Joules.
1. Identify the given data:
- Mass of water ([tex]\( m \)[/tex]): 100 grams
- Initial temperature of water ([tex]\( T_i \)[/tex]): 23.4°C
- Final temperature of water ([tex]\( T_f \)[/tex]): 20.9°C
- Specific heat capacity of water ([tex]\( c \)[/tex]): 4.18 J/g°C
2. Calculate the change in temperature ([tex]\( \Delta T \)[/tex]):
[tex]\[ \Delta T = T_i - T_f \][/tex]
Substituting the values:
[tex]\[ \Delta T = 23.4°C - 20.9°C = 2.5°C \][/tex]
3. Apply the formula to calculate the energy lost ([tex]\( Q \)[/tex]):
The formula to calculate the energy lost or gained is:
[tex]\[ Q = mc\Delta T \][/tex]
Where:
- [tex]\( Q \)[/tex] is the energy transferred,
- [tex]\( m \)[/tex] is the mass of the water,
- [tex]\( c \)[/tex] is the specific heat capacity,
- [tex]\( \Delta T \)[/tex] is the change in temperature.
4. Substitute the known values into the formula:
[tex]\[ Q = 100 \, \text{g} \times 4.18 \, \text{J/g°C} \times 2.5 \, \text{°C} \][/tex]
5. Perform the multiplication:
[tex]\[ Q = 100 \times 4.18 \times 2.5 = 1045.0 \, \text{J} \][/tex]
So, the energy lost by the water to the iron is 1045.0 Joules.
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