From everyday questions to specialized queries, IDNLearn.com has the answers. Find accurate and detailed answers to your questions from our experienced and dedicated community members.
Sagot :
Answer:
[tex]\( 30 \, \text{J/C} \)[/tex]
Explanation:
The heat lost by the hot water is equal to the heat gained by the cold water plus the heat absorbed by the calorimeter. The calorimeter constant represents the heat capacity of the calorimeter.
Solving:
Formula for hot water:
[tex]q_{\text{hot}} = m_{\text{hot}} \cdot c_{\text{water}} \cdot (T_{\text{initial, hot}} - T_{\text{final}})[/tex]
[tex]\hrulefill[/tex]
Given:
[tex]\item \( m_{\text{hot}} = 50 \, \text{ml} = 50 \, \text{g} \) (The density of water is \( 1 \, \text{g/ml}) \ \item \( c_{\text{water}} = 4.18 \, \text{J/g*C} \) (specific heat capacity of water) \item \( T_{\text{initial, hot}} = 85 \, \text{C} \) \item \( T_{\text{final}} = 53 \, \text{C} \)[/tex]
[tex]\hrulefill[/tex]
[tex]q_{\text{hot}} = 50 \, \text{g} \cdot 4.18 \, \text{J/gC} \cdot (85 \, \text{C} - 53 \, \text{C})\\ \\ q_{\text{hot}} = 50 \cdot 4.18 \cdot 32\\\\ \boxed{q_{\text{hot}} = 6692 \, \text{J}\\}[/tex]
[tex]\hrulefill[/tex]
Formula for cold water:
[tex]\[ q_{\text{cold}} = m_{\text{cold}} \cdot c_{\text{water}} \cdot (T_{\text{final}} - T_{\text{initial, cold}}) \][/tex]
[tex]\hrulefill[/tex]
Given:
[tex]\item \( m_{\text{cold}} = 50 \, \text{ml} = 50 \, \text{g} \) \item \( T_{\text{initial, cold}} = 25 \, \text{C} \)[/tex]
[tex]\hrulefill[/tex]
[tex]q_{\text{cold}} = 50 \, \text{g} \cdot 4.18 \, \text{J/gC} \cdot (53 \, \text{C} - 25 \, \text{C})\\ q_{\text{cold}} = 50 \cdot 4.18 \cdot 28\\ \boxed{q_{\text{cold}} = 5852 \, \text{J}}[/tex]
[tex]\hrulefill[/tex]
Calorimeter constant:
[tex]\[ C_{\text{cal}} = \frac{q_{\text{cal}}}{\Delta T} \] \\\\ Where \( \Delta T \) is the change in temperature of the water, which is the same for both hot and cold water: \[[/tex]
[tex]\hrulefill[/tex]
[tex]\Delta T = T_{\text{final}} - T_{\text{initial, cold}} = 53 \, \text{C} - 25 \, \text{C} = 28 \, \text{C} \\ C_{\text{cal}} = \frac{840 \, \text{J}}{28 \, \text{C}}\\ \boxed{ C_{\text{cal}} = 30 \, \text{J/C}}[/tex]
[tex]\hrulefill[/tex]
[tex]\text{Therefore, the calorimeter constant is \( 30 \, \text{J/C} \).}[/tex]
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. IDNLearn.com is your reliable source for answers. We appreciate your visit and look forward to assisting you again soon.