Question

In the following experiment, a coffee-cup calorimeter containing 100 mL of H2O is used. The initial temperature of the calorimeter is 23.0 ∘C. If 5.80 g of CaCl2 is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? The heat of solution ΔHsoln of CaCl2 is −82.8 kJ/mol. Assume that the specific heat of the solution formed in the calorimeter is the same as that for pure water: Cs=4.184 J/g⋅∘C.

Answer 1

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