IDNLearn.com makes it easy to find answers and share knowledge with others. Our platform provides detailed and accurate responses from experts, helping you navigate any topic with confidence.
Sagot :
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].
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Thanks for visiting IDNLearn.com. We’re dedicated to providing clear answers, so visit us again for more helpful information.