Answer :
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].
