• Lab apron
  • Eye protection
  • Centigram or analytical balance
  • 8-10g or unknown salt
  • Water
  • Thermometer
  • 3 Styrofoam cups (calorimeter)
  • 100 mL cylinder
  • Filter paper


  • Safety equipment, including goggles were put on.
  • 8-10g of an unknown salt on a filter paper was obtained from the teacher. The mass of the salt and filter paper were measured on the balance and was recorded.
  • The salt was taken to the work station. A calorimeter, thermometer, and a 100 mL cylinder were obtained and brought to the work station.
  • One hundred millilitres of water was measured and placed into the calorimeter. The lid was placed on the calorimeter.
  • The temperature of the water was then taken with the thermometer and recorded as the initial temperature of the water.
  • The unknown salt was poured into the calorimeter filled with water.
  • The lid was placed on the calorimeter. The thermometer was placed in the calorimeter for 5 minutes.
  • The temperature was recorded as the final temperature.
  • The lid was taken off the calorimeter and qualitative observations were recorded.
  • The mass of the filter paper was then recorded using the balance in order to later find the mass of the solid.
  • The materials were cleaned and the salt was properly disposed down the sink. The filter paper was properly disposed in the garbage.

Data Collection:

The following chart represents the quantitative observations and their measurements that were taken during the lab.

Type Quantity
m filter paper 0.80 g
m filter paper + salt 10.13 g
m salt 9.3 g
m water 100 g
c water 4.18 kJ/kg °C
t1 23.5 °C
t2 18° C
Δ t -5.5°C


Mass of Salt

m salt = (m filter paper + salt) – (mass filter paper)

m salt = (10.13 g) – (0.80 g)

m salt = 9.33 g

Change in Temperature

Δt = (t2) – (t1)

Δt = (18°C) – (23.5°C)

Δt = -5.5°C


1 mL = 1g

100 mL= 100 g


ΔHsystem= – surroundings

ΔH = -mcΔt

ΔH = (100g) (4.18 kJ/kg°C) (-5.5°C)

ΔH= 2299 J


1000 kJ = 1 J

ΔH in kJ = ΔH in J/1000

ΔH= 2299 J/1000

ΔH in kJ= 2.299

ΔHsol = (ΔH in kJ) / (msalt in g)

ΔHsol= (2.299 kJ) / (9.33 g)

ΔHsol= 0.2464094 kJ/g

ΔHsol= 0.246 kJ/g

After the calculations of the enthalpy were put together, it could be determined that the unknown salt was Potassium Chloride.

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