• 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.

author avatar
William Anderson (Schoolworkhelper Editorial Team)
William completed his Bachelor of Science and Master of Arts in 2013. He current serves as a lecturer, tutor and freelance writer. In his spare time, he enjoys reading, walking his dog and parasailing. Article last reviewed: 2022 | St. Rosemary Institution © 2010-2024 | Creative Commons 4.0

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