Calculate the specific heat of solution in Joules for dissolving 8.00 g of NH4NO3 in 100mL water by using the following equation: 0=qrxn + qsoln. Given: mass of calorimeter cup:2.96g Given: mass of water and calorimeter cup: 102.63g Given: mass of ammonium nitrate: 8.02g given : initial temperature of water :22.1 Celsius Given: Delta T: 6.1 Celsius Given: the coffe cup calorimeter absorbs no heat so qcal is negligible. Given: The specific heat of the solution is the same as pure water which is 4.184J/g Celsius. ÂÂ
Calculate the specific heat of solution in Joules for dissolving 8.00 g of NH4NO3 in 100mL water by using the following equation: 0=qrxn + qsoln.
Given: mass of calorimeter cup:2.96gÂÂ
Given: mass of water and calorimeter cup: 102.63g
Given: mass of ammonium nitrate: 8.02g
given : initial temperature of water :22.1 CelsiusÂÂ
Given: Delta T: 6.1 CelsiusÂÂ
Given: the coffe cup calorimeter absorbs no heat so qcal is negligible.ÂÂ
Given: The specific heat of the solution is the same as pure water which is 4.184J/g Celsius.
The Correct Answer and Explanation is :
To calculate the specific heat of the solution for dissolving ( \text{NH}_4\text{NO}_3 ), let’s break it down step by step.
Step 1: Understand the relationship
The total heat exchanged is:
[
q_{\text{rxn}} + q_{\text{soln}} = 0
]
Since ( q_{\text{rxn}} ) is the heat released or absorbed during the reaction, it is equal in magnitude but opposite in sign to ( q_{\text{soln}} ). Thus:
[
q_{\text{rxn}} = -q_{\text{soln}}
]
Step 2: Define ( q_{\text{soln}} )
[
q_{\text{soln}} = m_{\text{soln}} \cdot c_{\text{soln}} \cdot \Delta T
]
Where:
- ( m_{\text{soln}} ): Total mass of the solution (water + ( \text{NH}_4\text{NO}_3 ))
- ( c_{\text{soln}} ): Specific heat capacity (same as water, ( 4.184 \, \text{J/g}^\circ\text{C} ))
- ( \Delta T ): Change in temperature (( 6.1^\circ\text{C} ))
Step 3: Calculate total mass of solution
The total mass is:
[
m_{\text{soln}} = \text{mass of water} + \text{mass of NH}4\text{NO}_3 ] Given: [ \text{Mass of water} = \text{Mass of water and calorimeter cup} – \text{Mass of calorimeter cup} = 102.63 \, \text{g} – 2.96 \, \text{g} = 99.67 \, \text{g} ] [ m{\text{soln}} = 99.67 \, \text{g} + 8.02 \, \text{g} = 107.69 \, \text{g}
]
Step 4: Calculate ( q_{\text{soln}} )
Substitute values into the equation:
[
q_{\text{soln}} = 107.69 \, \text{g} \cdot 4.184 \, \frac{\text{J}}{\text{g}^\circ\text{C}} \cdot 6.1^\circ\text{C}
]
[
q_{\text{soln}} = 2744.2 \, \text{J}
]
Step 5: Calculate ( q_{\text{rxn}} )
[
q_{\text{rxn}} = -q_{\text{soln}}
]
[
q_{\text{rxn}} = -2744.2 \, \text{J}
]
Explanation
- The solution’s specific heat was assumed to be ( 4.184 \, \text{J/g}^\circ\text{C} ), identical to water, per the problem statement.
- The total mass was calculated by adding the mass of water and ( \text{NH}_4\text{NO}_3 ).
- Using the formula ( q = mc\Delta T ), we calculated the heat absorbed by the solution, which is positive because the solution absorbs heat.
- The reaction releases the same amount of heat but with a negative sign.
Hence, the specific heat of the solution corresponds to the energy exchanged during the dissolution, which is ( q_{\text{soln}} = 2744.2 \, \text{J} ).