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 = 750 gT1​ = 12 °C → T2​ = 45 °Cq = ?

How much heat q is needed to warm 750 g of water from 12 °C to 45 °C? The specific heat of water is 4.18 J/(g·°C). Give the answer in kJ.

m = 750 gT1 = 12 °CT2 = 45 °Cc = 4.18 J/(g·°C)
q = kJ
2.Easy
waterm = 100 gT1​ = 16 °C → T2​ = 50 °Cq = ?

How much heat q is needed to warm 100 g of water from 16 °C to 50 °C? The specific heat of water is 4.18 J/(g·°C). Give the answer in kJ.

m = 100 gT1 = 16 °CT2 = 50 °Cc = 4.18 J/(g·°C)
q = kJ
3.Medium
Tf​ = ?Aluminiumm = 200 gT = 110 °Cc = 0.897 J/(g·°C)Waterm = 150 gT = 15 °Cc = 4.18 J/(g·°C)

A 200 g block of aluminium (c = 0.897 J/(g·°C)) at 110 °C is dropped into 150 g of water (c = 4.18 J/(g·°C)) at 15 °C in an insulated cup. Find the final temperature Tf.

mmetal = 200 gcmetal = 0.897 J/(g·°C)Tmetal = 110 °Cmwater = 150 gTwater = 15 °Ccwater = 4.18 J/(g·°C)
Tf = °C
4.Medium
Tf​ = ?Copperm = 150 gT = 160 °Cc = 0.385 J/(g·°C)Waterm = 150 gT = 17 °Cc = 4.18 J/(g·°C)

A 150 g block of copper (c = 0.385 J/(g·°C)) at 160 °C is dropped into 150 g of water (c = 4.18 J/(g·°C)) at 17 °C in an insulated cup. Find the final temperature Tf.

mmetal = 150 gcmetal = 0.385 J/(g·°C)Tmetal = 160 °Cmwater = 150 gTwater = 17 °Ccwater = 4.18 J/(g·°C)
Tf = °C
5.Hard
H20 g C2​H5​OHC2​H5​OH + 3O₂reactants2CO₂ + 3H₂O(l)productsq = ?

Use the standard enthalpies of formation to find the heat q released when 20 g of C2H5OH (M = 46.07 g/mol) burns completely: C2H5OH + 3O₂ → 2CO₂ + 3H₂O(l). Give q as a positive number.

ΔH°f(C2H5OH) = −277.7 kJ/molΔH°f(CO₂) = −393.5 kJ/molΔH°f(H₂O, l) = −285.8 kJ/molΔH°f(O₂) = 0
q = kJ
6.Hard
H2C4​H10​ + 13O₂reactants8CO₂ + 10H₂O(l)productsΔH° = ?

Use the standard enthalpies of formation to find ΔH° for the reaction 2C4H10 + 13O₂ → 8CO₂ + 10H₂O(l).

ΔH°f(C4H10) = −125.6 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 = 103.5 kJ
  2. 2. q = 14.21 kJ
  3. 3. Tf = 36.13 °C
  4. 4. Tf = 29.06 °C
  5. 5. q = 593.3 kJ
  6. 6. ΔH° = −5755 kJ

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