How much energy in kJ is released when 17.55 g of water at 47.5 °C is cooled to –33.5 °C?
M.P. = 0 °C; B.P. = 100 °C; All specific heats in J/(g*K): s(ice) = 2.03; s(water) = 4.184; s(steam) = 1.890. heat of fusion = 6.01 kJ/mol; heat of vaporization = 40.67 kJ/mol.
q1 = heat released to cool from 47.5 to zero C.
q2 = heat released at 0 C to freeze water @ zero C.
q3 = heat released to cool from zero C to -33.5 C.
Total is q1 + q2 + q3
q1 = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q2 = mass H2O x heat fusion.
q3 = mass ice x specific heat ice x (Tfinal-Tinitial)
Note: q1 will be in J if you use 4.184 for sp. heat.
q2 will be in kJ AND not it is in kJ/mol; therefore, you must change that to kJ/g or change mass H2O from grams to mols.
q3 will be in J if you use 2.03.
finally, q1, q2, q3 must be in J or kJ to add them.
Post your work if you get stuck.
To calculate the energy released when cooling water from 47.5 °C to -33.5 °C, we need to consider the different phases of water and the energy changes associated with each phase transition.
First, we need to calculate the energy required to cool the water from 47.5 °C to 0 °C.
The specific heat capacity of water (s(water)) is given as 4.184 J/(g*K). Therefore, we can use the formula:
q = m * s * ΔT
Where:
q is the energy absorbed or released (in J),
m is the mass of the water (in g),
s is the specific heat capacity of water (in J/(g*K)),
ΔT is the change in temperature (in °C).
Using the given values, we can calculate the amount of energy required to cool the water from 47.5 °C to 0 °C:
q1 = 17.55 g * 4.184 J/(g*K) * (0 °C - 47.5 °C)
= -3551.137 J
Next, we need to consider the phase change from 0 °C to -33.5 °C. During this phase change, the water will be in the solid phase (ice). The energy involved in this phase change is known as the heat of fusion.
The heat of fusion is given as 6.01 kJ/mol. We need to convert grams of water to moles to calculate the energy:
Step 1: Convert grams to moles using the molar mass of water (18.015 g/mol):
moles of water = 17.55 g / 18.015 g/mol ≈ 0.972 mol
Step 2: Calculate the energy involved in the phase change:
q2 = 0.972 mol * (6.01 kJ/mol)
= 5.82 kJ
Finally, we need to calculate the energy required to cool the water from 0 °C to -33.5 °C.
The specific heat capacity of ice (s(ice)) is given as 2.03 J/(g*K). Therefore, we can use the formula:
q3 = m * s * ΔT
Where:
q3 is the energy absorbed or released (in J),
m is the mass of the water (in g),
s is the specific heat capacity of ice (in J/(g*K)),
ΔT is the change in temperature (in °C).
Using the given values, we can calculate the amount of energy required to cool the water from 0 °C to -33.5 °C:
q4 = 17.55 g * 2.03 J/(g*K) * (-33.5 °C - 0 °C)
= -1189.5805 J
To calculate the total energy released, we sum up the energies involved at each step:
Total energy released = q1 + q2 + q3
= -3551.137 J + 5.82 kJ - 1189.5805 J
≈ -3654.8975 J
To convert this energy to kJ, we divide the value by 1000:
Total energy released ≈ -3.65 kJ
Therefore, approximately 3.65 kJ of energy is released when 17.55 g of water at 47.5 °C is cooled to -33.5 °C.