Heat It Up
Capacity for Change
Calorimetry Corner
Reaction Reaction
Phase Change Parade
100

This is the ability to do work and transfer heat. What is the unit?

Energy (J)

100

The amount of energy required to raise 1 gram of a substance 1°C is called this.

Specific heat capacity

100

50 g of water at 20°C absorbs 4180 J of heat. What is its final temperature? (c_water = 4.18 J/g°C)

ΔT = Q / (m·c) = 4180 / (50×4.18) ≈ 20°C → T_final = 20 + 20 = 40°C

100

ΔH = +50 kJ for the reaction: A → B. Is the reaction endothermic or exothermic?

Endothermic

100

The energy required to melt a solid without changing its temperature is called this.

Heat of Fusion

200

True or False: During a phase change, the temperature of the substance increases

False

200

Calculate the heat absorbed by 50 g of water if it is heated from 25°C to 75°C. (Specific heat of water = 4.18 J/g°C)

Q = 50 × 4.18 × (75-25) = 10,450 J

200

A 25 g metal sample at 90°C is placed in 100 g of water at 25°C. The final temperature is 30°C. If c_water = 4.18 J/g°C, calculate the specific heat of the metal.

q_water = q_metal → 100×4.18×(30-25) = 25×c×(90-30) → 2090 = 25×c×60 → c ≈ 1.39 J/g°C

200

Unbalanced reaction: NH₄NO₃(s) → N₂O(g) + H₂O(g) ΔH = -36 kJ
If 56 g of NH₄NO₃ decomposes, how much energy is released?

56 g / 80 g/mol = 0.7 mol → q = 0.7 × -36 kJ = -25.2 kJ

200

The energy needed to vaporize 1 mole of a liquid at its boiling point is called this.

Heat of vaporization

300

How does heat transfure?

From Hot to Cold

300

If 2500 J of heat raises the temperature of 100 g of an unknown metal by 10°C, its specific heat is this.

c = 2500 / (100 × 10) = 2.5 J/g°C

300

A 55.0 g unknown metal at 75°C is placed in a calorimeter containing 50.0 g of water at 25°C. The final temperature is 35°C. Calculate the specific heat of the metal. (c_water = 4.18 J/g°C)
 

q_water = m·c·ΔT = 50×4.18×(35-25) = 50×4.18×10 = 2090 J
q_metal = m·c·ΔT = 55×c×(75-35) = 55×c×40
q_metal = q_water → 55×c×40 = 2090 → c ≈ 1.20 J/g°C

300

Unbalanced reaction: NH₄NO₃(s) → N₂O(g) + H₂O(g) ΔH = -36 kJ 

Using the same reaction, how much NH₄NO₃ (in grams) is needed to release 170 kJ of heat?

170 kJ / 36 kJ/mol = 4.72 mol → mass = 4.72×80 ≈ 378 g

300

The heat of vaporization is equal in magnitude to?

Heat of condensation

400

This describes a process in which heat is moving from the system to the surroundings. (What is the sign)

Exothermic (sign qsys = -)

400

What is the change in temperature when 2000 J of energy is given to 40 g of iron? (c = 0.45 J/g°C) 

q = (40 g)(2000 J)(0.45 J/g°C) = 111 °C

400

80 g of an unknown metal at 100°C is dropped into 200 g of water at 25°C. Final temperature is 30°C. Find the specific heat of the metal. (c_water = 4.18 J/g°C)

q_water = 200×4.18×(30-25) = 4180 J

400

The reaction: 2 Al + 3 Cl₂ → 2 AlCl₃ ΔH = -704 kJ. How much energy is released when 54 g of Al reacts?

54 g / 27 g/mol = 2 mol → reaction releases 2/2 × -704 = -704 kJ

400

10 g of Ice at -5°C is heated to water at 25°C. This step requires calculating heat for heating the solid, melting, and heating the liquid. The term for this total energy is?


Enthalpies and Specific Heat of Water

Heat of Fusion ΔHfus

334 J/g

Heat of Vaporization ΔHvap

2260 J/g

Heat capacity H2O (s)

2.11 J/gºC

Heat capacity H2O (l)

4.18 J/gºC

Heat capacity H2O (g)

2.08 J/gºC

 

4.49 kJ or 4490 J

500

This describes a process in which heat is moving from the surroundings to the system. (What is the sign)

Endothermic (sign qsys = +)

500

How much heat is needed to raise 75 g of iron (c = 0.45 J/g°C) from 20°C to 95°C?
 

Q = 75 × 0.45 × (95-20) = 2,531.25 J

500

A 40 g metal at 95°C is placed in 150 g water at 20°C. The final temperature is 25°C. If ΔH for water is negative, what is the energy absorbed by the water, and the specific heat of the metal?

q_water = 150×4.18×(25-20) = 150×4.18×5 = 3135 J
q_metal = 40×c×(95-25) = 40×c×70
40×c×70 = 3135 → c ≈ 1.12 J/g°C

500

The combustion of CH₄: CH₄ + 2 O₂ → CO₂ + 2 H₂O ΔH = -890 kJ/mol. How much heat is released when 16 g of CH₄ burns?

16 g / 16 g/mol = 1 mol → q = -890 kJ

500

Calculate the total heat required to turn 10 g of ice at -10°C into steam at 110°C. 


Enthalpies and Specific Heat of Water

Heat of Fusion ΔHfus

334 J/g

Heat of Vaporization ΔHvap

2260 J/g

Heat capacity H2O (s)

2.11 J/gºC

Heat capacity H2O (l)

4.18 J/gºC

Heat capacity H2O (g)

2.08 J/gºC


Answer:

  1. Heat ice from -10 → 0°C: 10 × 2.09 × 10 = 209 J
  2. Melt ice: 10 × 334 = 3340 J
  3. Heat water 0 → 100°C: 10 × 4.18 × 100 = 4180 J
  4. Vaporize water: 10 × 2260 = 22,600 J
  5. Heat steam 100 → 110°C: 10 × 2 × 10 = 200 J
    Total Q = 209 + 3340 + 4180 + 22,600 + 200 = 30,529 J