Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
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 appreciate your time. Please revisit us for more reliable answers to any questions you may have. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.