Answer :
To find out how much energy the water absorbed in the experiment according to the calorimeter data, we'll follow these steps:
1. Determine the given values from the calorimeter data:
- Mass of the water, [tex]\( m = 100.0 \, \text{g} \)[/tex]
- Specific heat of water, [tex]\( c = 4.18 \, \text{J/g}^\circ\text{C} \)[/tex]
- Initial temperature, [tex]\( T_i = 21.2^\circ\text{C} \)[/tex]
- Final temperature, [tex]\( T_f = 46.2^\circ\text{C} \)[/tex]
2. Calculate the change in temperature [tex]\(\Delta T\)[/tex]:
[tex]\[ \Delta T = T_f - T_i \][/tex]
Substituting the values:
[tex]\[ \Delta T = 46.2^\circ\text{C} - 21.2^\circ\text{C} = 25.0^\circ\text{C} \][/tex]
3. Use the formula for the amount of heat absorbed or released in a calorimetry process:
[tex]\[ q = m \cdot c \cdot \Delta T \][/tex]
where [tex]\( q \)[/tex] is the heat absorbed, [tex]\( m \)[/tex] is the mass, [tex]\( c \)[/tex] is the specific heat, and [tex]\(\Delta T\)[/tex] is the change in temperature.
4. Substitute the values into the formula:
[tex]\[ q = 100.0 \, \text{g} \times 4.18 \, \text{J/g}^\circ\text{C} \times 25.0^\circ\text{C} \][/tex]
5. Calculate the energy absorbed:
[tex]\[ q = 100.0 \times 4.18 \times 25.0 = 10450.0 \, \text{J} \][/tex]
Therefore, the energy the water absorbed in this experiment is [tex]\( 10450.0 \, \text{J} \)[/tex].
1. Determine the given values from the calorimeter data:
- Mass of the water, [tex]\( m = 100.0 \, \text{g} \)[/tex]
- Specific heat of water, [tex]\( c = 4.18 \, \text{J/g}^\circ\text{C} \)[/tex]
- Initial temperature, [tex]\( T_i = 21.2^\circ\text{C} \)[/tex]
- Final temperature, [tex]\( T_f = 46.2^\circ\text{C} \)[/tex]
2. Calculate the change in temperature [tex]\(\Delta T\)[/tex]:
[tex]\[ \Delta T = T_f - T_i \][/tex]
Substituting the values:
[tex]\[ \Delta T = 46.2^\circ\text{C} - 21.2^\circ\text{C} = 25.0^\circ\text{C} \][/tex]
3. Use the formula for the amount of heat absorbed or released in a calorimetry process:
[tex]\[ q = m \cdot c \cdot \Delta T \][/tex]
where [tex]\( q \)[/tex] is the heat absorbed, [tex]\( m \)[/tex] is the mass, [tex]\( c \)[/tex] is the specific heat, and [tex]\(\Delta T\)[/tex] is the change in temperature.
4. Substitute the values into the formula:
[tex]\[ q = 100.0 \, \text{g} \times 4.18 \, \text{J/g}^\circ\text{C} \times 25.0^\circ\text{C} \][/tex]
5. Calculate the energy absorbed:
[tex]\[ q = 100.0 \times 4.18 \times 25.0 = 10450.0 \, \text{J} \][/tex]
Therefore, the energy the water absorbed in this experiment is [tex]\( 10450.0 \, \text{J} \)[/tex].
