Calorimetry & enthalpy worksheet with answers

Free printable Calorimetry & enthalpy worksheet: 6 problems from easy to exam level, with an answer key. Print it or save it as a PDF.

New worksheetPractise online with hints

For teachers: free to print, copy and use in class. Every new worksheet has new values and its own answer key. A link to ExamProblems on your course page is appreciated.

Name: ____________________   Date: ____________

1.Easy
waterm = 150 gT1​ = 21 °C → T2​ = 45 °Cq = ?

How much heat q is needed to warm 150 g of water from 21 °C to 45 °C? The specific heat of water is 4.18 J/(g·°C). Give the answer in kJ.

m = 150 gT1 = 21 °CT2 = 45 °Cc = 4.18 J/(g·°C)
q = kJ
2.Easy
waterm = 150 gq = 11 kJΔT = ?

11 kJ of heat is added to 150 g of water (c = 4.18 J/(g·°C)). Find the temperature rise ΔT.

q = 11 kJm = 150 gc = 4.18 J/(g·°C)
ΔT = °C
3.Medium
Tf​ = ?Ironm = 75 gT = 120 °Cc = 0.449 J/(g·°C)Waterm = 100 gT = 20 °Cc = 4.18 J/(g·°C)

A 75 g block of iron (c = 0.449 J/(g·°C)) at 120 °C is dropped into 100 g of water (c = 4.18 J/(g·°C)) at 20 °C in an insulated cup. Find the final temperature Tf.

mmetal = 75 gcmetal = 0.449 J/(g·°C)Tmetal = 120 °Cmwater = 100 gTwater = 20 °Ccwater = 4.18 J/(g·°C)
Tf = °C
4.Medium
Tf​ = ?Aluminiumm = 100 gT = 150 °Cc = 0.897 J/(g·°C)Waterm = 150 gT = 24 °Cc = 4.18 J/(g·°C)

A 100 g block of aluminium (c = 0.897 J/(g·°C)) at 150 °C is dropped into 150 g of water (c = 4.18 J/(g·°C)) at 24 °C in an insulated cup. Find the final temperature Tf.

mmetal = 100 gcmetal = 0.897 J/(g·°C)Tmetal = 150 °Cmwater = 150 gTwater = 24 °Ccwater = 4.18 J/(g·°C)
Tf = °C
5.Hard
H250 g C6​H12​O6​C6​H12​O6​ + 6O₂reactants6CO₂ + 6H₂O(l)productsq = ?

Use the standard enthalpies of formation to find the heat q released when 250 g of C6H12O6 (M = 180.16 g/mol) burns completely: C6H12O6 + 6O₂ → 6CO₂ + 6H₂O(l). Give q as a positive number.

ΔH°f(C6H12O6) = −1274 kJ/molΔH°f(CO₂) = −393.5 kJ/molΔH°f(H₂O, l) = −285.8 kJ/molΔH°f(O₂) = 0
q = kJ
6.Hard
HC6​H12​O6​ + 6O₂reactants6CO₂ + 6H₂O(l)productsΔH° = ?

Use the standard enthalpies of formation to find ΔH° for the reaction C6H12O6 + 6O₂ → 6CO₂ + 6H₂O(l).

ΔH°f(C6H12O6) = −1274 kJ/molΔH°f(CO₂) = −393.5 kJ/molΔH°f(H₂O, l) = −285.8 kJ/molΔH°f(O₂) = 0
ΔH° = kJ

Answer key

  1. 1. q = 15.05 kJ
  2. 2. ΔT = 17.54 °C
  3. 3. Tf = 27.46 °C
  4. 4. Tf = 39.77 °C
  5. 5. q = 3888 kJ
  6. 6. ΔH° = −2802 kJ

Full step-by-step solutions: practise this topic on examproblems.com.