Answer:
The final temperature is 32.78°C
Explanation:
Step 1: Data given
Volume of water = 100 mL = 0.100 L
The initial temperature of the calorimeter is 23.0 ∘C.
Mass of CaCl2 = 5.80 grams
The heat of solution ΔHsoln of CaCl2 is −82.8 kJ/mol
Specific heat of calorimeter = 4.184 J/g°C
Step 2: Calculate moles CaCl2
Moles CaCl2 = mass CaCl2 / molar mass CaCl2
Moles CaCl2 = 5.80 grams / 110.98 g/mol
Moles CaCl2 = 0.0523 moles
Step 3: Calculate energy
Q = ΔH * moles
Q = 82.8 kJ/mol * 0.0523 moles
Q = 4.33 kJ
Step 4: Calculate change of temperature
Q = m*C*ΔT
⇒with Q = the energy needed = 4.33 kJ
⇒with m = the mass = 105.80 grams
⇒with C = the specific heat = 4.184 J/g°C
⇒with ΔT = the change of temperature = TO BE DETERMINED
ΔT = 4.330 J / (105.80 * 4.184 J/g°C)
ΔT = 9.78 °C
Step 5: Calculate the final temperature
ΔT = T2 - T1
9.78 °C = T2-23.0°C
T2 = 32.78 °C
The final temperature is 32.78°C
