Discover the answers you need at Westonci.ca, where experts provide clear and concise information on various topics. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
Let's walk through the steps for solving the problem using the given data:
1. Mass of water ([tex]\(m_{\text{water}}\)[/tex]): 50.00 g
2. Initial temperature of water ([tex]\(T_{\text{initial}}\)[/tex]): [tex]\(50^\circ C\)[/tex]
3. Final temperature of water ([tex]\(T_{\text{final}}\)[/tex]): [tex]\(2^\circ C\)[/tex]
4. Mass of ice ([tex]\(m_{\text{ice}}\)[/tex]): 900 g
5. Specific heat capacity of water/ice ([tex]\(C\)[/tex]): [tex]\(4.184 \frac{J}{g \cdot C}\)[/tex]
### Part a) Heat absorbed by the ice and its sign
The heat absorbed or released by a substance can be calculated using the formula:
[tex]\[ Q = mc\Delta T \][/tex]
where:
- [tex]\(Q\)[/tex] is the heat absorbed or released (in Joules)
- [tex]\(m\)[/tex] is the mass (in grams)
- [tex]\(c\)[/tex] is the specific heat capacity (in [tex]\( \frac{J}{g \cdot C} \)[/tex])
- [tex]\(\Delta T\)[/tex] is the change in temperature (final temperature - initial temperature)
For the water, we need to calculate [tex]\(\Delta T\)[/tex]:
[tex]\[ \Delta T_{\text{water}} = T_{\text{final}} - T_{\text{initial}} = 2^\circ C - 50^\circ C = -48^\circ C \][/tex]
Now let's use the heat formula:
[tex]\[ Q_{\text{water}} = m_{\text{water}} \times C \times \Delta T_{\text{water}} \][/tex]
[tex]\[ Q_{\text{water}} = 50.00 \, \text{g} \times 4.184 \, \frac{J}{g \cdot C} \times (-48^\circ C) \][/tex]
[tex]\[ Q_{\text{water}} = -10041.6 \, \text{J} \][/tex]
Therefore, the heat absorbed by the ice is [tex]\(-10041.6 \, \text{J}\)[/tex]. Since the value is negative, it indicates that the heat is released by the water, making it exothermic.
### Part b) Heat capacity of the ice
To calculate the heat capacity ([tex]\(C_{\text{ice}}\)[/tex]) of the ice, we will use the same heat formula but rearrange it to solve for [tex]\(C_{\text{ice}}\)[/tex]:
[tex]\[ Q_{\text{ice}} = m_{\text{ice}} \times C_{\text{ice}} \times \Delta T_{\text{ice}} \][/tex]
Given that [tex]\(\Delta T_{\text{ice}} = 30^\circ C\)[/tex] and knowing the heat absorbed/released by the ice is [tex]\(Q_{\text{ice}} = -10041.6 \, \text{J}\)[/tex]:
[tex]\[ -10041.6 \, \text{J} = 900 \, \text{g} \times C_{\text{ice}} \times 30^\circ C \][/tex]
Solving for [tex]\(C_{\text{ice}}\)[/tex]:
[tex]\[ C_{\text{ice}} = \frac{-10041.6 \, \text{J}}{900 \, \text{g} \times 30^\circ C} \][/tex]
[tex]\[ C_{\text{ice}} = \frac{-10041.6}{27000} \][/tex]
[tex]\[ C_{\text{ice}} = -0.372 \, \text{J}/^\circ C \][/tex]
Calculating value of C:
[tex]\[ C_{\text{ice}} = \frac{-10041.6}{30 \, \text{C}} \][/tex]
[tex]\[ C_{\text{ice}} = -334.72 \, \text{J}/^\circ C \][/tex]
### Part c) Percentage error in the calculation of part B
To find the percentage error compared to the given heat capacity of ice, we assume the heat capacity is the same as water, [tex]\(C = 4.184 \frac{J}{g \cdot C}\)[/tex].
First, compute the given heat capacity of ice:
[tex]\[ C_{\text{ice, given}} = m_{\text{ice}} \times C \][/tex]
[tex]\[ C_{\text{ice, given}} = 900 \, \text{g} \times 4.184 \, \frac{J}{g \cdot C} \][/tex]
[tex]\[ C_{\text{ice, given}} = 3765.6 \, \text{J}/^\circ C \][/tex]
Now, calculate the percentage error:
[tex]\[ \text{Percentage error} = \frac{\left| C_{\text{ice, given}} - C_{\text{ice}} \right|}{C_{\text{ice, given}}} \times 100 \][/tex]
[tex]\[ \text{Percentage error} = \frac{\left| 3765.6 - (-334.72) \right|}{3765.6} \times 100 \][/tex]
[tex]\[ \text{Percentage error} = \frac{4100.32}{3765.6} \times 100 \][/tex]
[tex]\[ \text{Percentage error} \approx 108.89\% \][/tex]
To summarize:
a) The heat absorbed by the ice is [tex]\(-10041.6 \, \text{J}\)[/tex], indicating that heat is released, so the heat is negative.
b) The heat capacity of the ice given the temperature change is [tex]\( -334.72 \, \text{J}/^\circ C \)[/tex].
c) The percentage error in the calculation of part B is approximately [tex]\(108.89\% \)[/tex].
1. Mass of water ([tex]\(m_{\text{water}}\)[/tex]): 50.00 g
2. Initial temperature of water ([tex]\(T_{\text{initial}}\)[/tex]): [tex]\(50^\circ C\)[/tex]
3. Final temperature of water ([tex]\(T_{\text{final}}\)[/tex]): [tex]\(2^\circ C\)[/tex]
4. Mass of ice ([tex]\(m_{\text{ice}}\)[/tex]): 900 g
5. Specific heat capacity of water/ice ([tex]\(C\)[/tex]): [tex]\(4.184 \frac{J}{g \cdot C}\)[/tex]
### Part a) Heat absorbed by the ice and its sign
The heat absorbed or released by a substance can be calculated using the formula:
[tex]\[ Q = mc\Delta T \][/tex]
where:
- [tex]\(Q\)[/tex] is the heat absorbed or released (in Joules)
- [tex]\(m\)[/tex] is the mass (in grams)
- [tex]\(c\)[/tex] is the specific heat capacity (in [tex]\( \frac{J}{g \cdot C} \)[/tex])
- [tex]\(\Delta T\)[/tex] is the change in temperature (final temperature - initial temperature)
For the water, we need to calculate [tex]\(\Delta T\)[/tex]:
[tex]\[ \Delta T_{\text{water}} = T_{\text{final}} - T_{\text{initial}} = 2^\circ C - 50^\circ C = -48^\circ C \][/tex]
Now let's use the heat formula:
[tex]\[ Q_{\text{water}} = m_{\text{water}} \times C \times \Delta T_{\text{water}} \][/tex]
[tex]\[ Q_{\text{water}} = 50.00 \, \text{g} \times 4.184 \, \frac{J}{g \cdot C} \times (-48^\circ C) \][/tex]
[tex]\[ Q_{\text{water}} = -10041.6 \, \text{J} \][/tex]
Therefore, the heat absorbed by the ice is [tex]\(-10041.6 \, \text{J}\)[/tex]. Since the value is negative, it indicates that the heat is released by the water, making it exothermic.
### Part b) Heat capacity of the ice
To calculate the heat capacity ([tex]\(C_{\text{ice}}\)[/tex]) of the ice, we will use the same heat formula but rearrange it to solve for [tex]\(C_{\text{ice}}\)[/tex]:
[tex]\[ Q_{\text{ice}} = m_{\text{ice}} \times C_{\text{ice}} \times \Delta T_{\text{ice}} \][/tex]
Given that [tex]\(\Delta T_{\text{ice}} = 30^\circ C\)[/tex] and knowing the heat absorbed/released by the ice is [tex]\(Q_{\text{ice}} = -10041.6 \, \text{J}\)[/tex]:
[tex]\[ -10041.6 \, \text{J} = 900 \, \text{g} \times C_{\text{ice}} \times 30^\circ C \][/tex]
Solving for [tex]\(C_{\text{ice}}\)[/tex]:
[tex]\[ C_{\text{ice}} = \frac{-10041.6 \, \text{J}}{900 \, \text{g} \times 30^\circ C} \][/tex]
[tex]\[ C_{\text{ice}} = \frac{-10041.6}{27000} \][/tex]
[tex]\[ C_{\text{ice}} = -0.372 \, \text{J}/^\circ C \][/tex]
Calculating value of C:
[tex]\[ C_{\text{ice}} = \frac{-10041.6}{30 \, \text{C}} \][/tex]
[tex]\[ C_{\text{ice}} = -334.72 \, \text{J}/^\circ C \][/tex]
### Part c) Percentage error in the calculation of part B
To find the percentage error compared to the given heat capacity of ice, we assume the heat capacity is the same as water, [tex]\(C = 4.184 \frac{J}{g \cdot C}\)[/tex].
First, compute the given heat capacity of ice:
[tex]\[ C_{\text{ice, given}} = m_{\text{ice}} \times C \][/tex]
[tex]\[ C_{\text{ice, given}} = 900 \, \text{g} \times 4.184 \, \frac{J}{g \cdot C} \][/tex]
[tex]\[ C_{\text{ice, given}} = 3765.6 \, \text{J}/^\circ C \][/tex]
Now, calculate the percentage error:
[tex]\[ \text{Percentage error} = \frac{\left| C_{\text{ice, given}} - C_{\text{ice}} \right|}{C_{\text{ice, given}}} \times 100 \][/tex]
[tex]\[ \text{Percentage error} = \frac{\left| 3765.6 - (-334.72) \right|}{3765.6} \times 100 \][/tex]
[tex]\[ \text{Percentage error} = \frac{4100.32}{3765.6} \times 100 \][/tex]
[tex]\[ \text{Percentage error} \approx 108.89\% \][/tex]
To summarize:
a) The heat absorbed by the ice is [tex]\(-10041.6 \, \text{J}\)[/tex], indicating that heat is released, so the heat is negative.
b) The heat capacity of the ice given the temperature change is [tex]\( -334.72 \, \text{J}/^\circ C \)[/tex].
c) The percentage error in the calculation of part B is approximately [tex]\(108.89\% \)[/tex].
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
