The molar heat of solution (ΔHsoln) of sodium hydroxide is -445.1 kJ/mol. In a certain experiment, 5.00 g of solid sodium hydroxide is completely dissolved in 1.000 L of 20.0 °C water in a foam cup calorimeter. Assuming no heat loss, calculate the final temperature of the water.
The Correct Answer and Explanation is:
To calculate the final temperature of the water when sodium hydroxide dissolves, we can use the heat energy equation and the molar heat of solution. Here’s a step-by-step approach:
Step 1: Calculate the moles of sodium hydroxide (NaOH)
The molar mass of sodium hydroxide (NaOH) is approximately 40.0 g/mol.moles of NaOH=mass of NaOHmolar mass of NaOH=5.00 g40.0 g/mol=0.125 mol\text{moles of NaOH} = \frac{\text{mass of NaOH}}{\text{molar mass of NaOH}} = \frac{5.00 \, \text{g}}{40.0 \, \text{g/mol}} = 0.125 \, \text{mol}moles of NaOH=molar mass of NaOHmass of NaOH=40.0g/mol5.00g=0.125mol
Step 2: Calculate the heat released during dissolution
The molar heat of solution (ΔH_soln) of sodium hydroxide is -445.1 kJ/mol, which means that for every mole of NaOH dissolved, 445.1 kJ of energy is released (exothermic process).Heat released=moles of NaOH×ΔHsoln=0.125 mol×(−445.1 kJ/mol)=−55.64 kJ\text{Heat released} = \text{moles of NaOH} \times \Delta H_{\text{soln}} = 0.125 \, \text{mol} \times (-445.1 \, \text{kJ/mol}) = -55.64 \, \text{kJ}Heat released=moles of NaOH×ΔHsoln=0.125mol×(−445.1kJ/mol)=−55.64kJ
Since the heat is released, we treat the heat energy as positive when it goes into the water.Qwater=55.64 kJQ_{\text{water}} = 55.64 \, \text{kJ}Qwater=55.64kJ
Step 3: Convert heat energy to joules
We need to convert kJ to joules, because the specific heat capacity of water is usually given in J/g°C.Qwater=55.64 kJ×1000 J/kJ=55640 JQ_{\text{water}} = 55.64 \, \text{kJ} \times 1000 \, \text{J/kJ} = 55640 \, \text{J}Qwater=55.64kJ×1000J/kJ=55640J
Step 4: Use the formula for heat transfer to calculate the temperature change
The formula for heat transfer is:Q=m⋅c⋅ΔTQ = m \cdot c \cdot \Delta TQ=m⋅c⋅ΔT
Where:
- QQQ is the heat energy (in joules),
- mmm is the mass of the water (in grams),
- ccc is the specific heat capacity of water (4.18 J/g°C),
- ΔT\Delta TΔT is the change in temperature (in °C).
For 1.000 L of water, the mass of the water is 1000 g (since 1 L of water is approximately 1000 g). Now, we can solve for the temperature change ΔT\Delta TΔT:55640 J=1000 g×4.18 J/g°C×ΔT55640 \, \text{J} = 1000 \, \text{g} \times 4.18 \, \text{J/g°C} \times \Delta T55640J=1000g×4.18J/g°C×ΔTΔT=55640 J1000 g×4.18 J/g°C=13.3 °C\Delta T = \frac{55640 \, \text{J}}{1000 \, \text{g} \times 4.18 \, \text{J/g°C}} = 13.3 \, \text{°C}ΔT=1000g×4.18J/g°C55640J=13.3°C
Step 5: Calculate the final temperature
The initial temperature of the water is 20.0°C. The final temperature will be the initial temperature plus the temperature change:Tfinal=20.0°C+13.3°C=33.3°CT_{\text{final}} = 20.0°C + 13.3°C = 33.3°CTfinal=20.0°C+13.3°C=33.3°C
Final Answer:
The final temperature of the water is 33.3°C.
Explanation:
The dissolution of sodium hydroxide is an exothermic process, meaning it releases heat into the water. Since the heat released is transferred to the water, the water temperature increases. By using the molar heat of solution, we calculated the heat released during the dissolution process. We then used the specific heat capacity of water to determine how much the temperature of the water would increase, leading to the final temperature of 33.3°C.
