Section A Measurement & Significant Figures - Answers
Problem 1
Record the volume of water in the graduated cylinder to the correct number of significant figures.
Answer: 36.80 mL (or 36.8 mL)
Explanation: When reading a graduated cylinder:
- Read from the bottom of the meniscus (curved surface of the liquid)
- The water level is at approximately 36.8 on the scale
- The markings show increments of 0.1 mL (every small line is 0.1 mL)
- When markings are every 0.1 mL, you should estimate to 0.01 mL (one place beyond the markings)
- The reading appears to be exactly on the 36.8 line, so we record it as 36.80 mL (4 sig figs)
- If recorded as 36.8 mL, this would be 3 sig figs (also acceptable depending on the precision you can observe)
Problem 2
How many significant figures in the following values?
a. 35.0 cm = 3 sig figs
b. 0.089 mL = 2 sig figs
c. 273.00°C = 5 sig figs
d. \(1.0890 \times 10^{-6}\) nm = 5 sig figs
b. 0.089 mL = 2 sig figs
c. 273.00°C = 5 sig figs
d. \(1.0890 \times 10^{-6}\) nm = 5 sig figs
Explanation:
a. All digits are significant, including the trailing zero after the decimal
b. Leading zeros are NOT significant; only 8 and 9 count
c. All digits are significant, including trailing zeros after the decimal
d. In scientific notation, all digits in the coefficient are significant (1, 0, 8, 9, 0)
a. All digits are significant, including the trailing zero after the decimal
b. Leading zeros are NOT significant; only 8 and 9 count
c. All digits are significant, including trailing zeros after the decimal
d. In scientific notation, all digits in the coefficient are significant (1, 0, 8, 9, 0)
Problem 3
Round the answers to the following to the correct number of significant figures:
a. \(35.00 \times 1.4 = \) 49
b. \(23.506 \div 21.00 = \) 1.119
c. \(23.016 + 10. = \) 33
d. \(109.00 - 57.9145 = \) 51.09
b. \(23.506 \div 21.00 = \) 1.119
c. \(23.016 + 10. = \) 33
d. \(109.00 - 57.9145 = \) 51.09
Work:
a. \(35.00 \times 1.4 = 49.0\) → Round to 2 sig figs (least in 1.4) = 49
b. \(23.506 \div 21.00 = 1.119238...\) → Round to 4 sig figs (least in both) = 1.119
c. \(23.016 + 10. = 33.016\) → Round to ones place (least precise in 10.) = 33
d. \(109.00 - 57.9145 = 51.0855\) → Round to hundredths place (least precise in 109.00) = 51.09
a. \(35.00 \times 1.4 = 49.0\) → Round to 2 sig figs (least in 1.4) = 49
b. \(23.506 \div 21.00 = 1.119238...\) → Round to 4 sig figs (least in both) = 1.119
c. \(23.016 + 10. = 33.016\) → Round to ones place (least precise in 10.) = 33
d. \(109.00 - 57.9145 = 51.0855\) → Round to hundredths place (least precise in 109.00) = 51.09
Section B Conceptual Understanding - Answers
Problem 4
What is the difference between mass and volume?
Mass is the amount of matter in an object (measured in grams). Volume is the amount of space an object occupies (measured in mL or cm³).
Key Differences:
- Mass: How much "stuff" is in an object; measured with a balance; units: g, kg
- Volume: How much space it takes up; measured with graduated cylinder or ruler; units: mL, L, cm³
- Objects can have the same volume but different masses (e.g., 1 L of water vs. 1 L of oil)
Section C Density Calculations - Answers
Problem 5
A block of wood weighing 15.0 g measures 2.4 cm on each side. What is its density? Will it sink or float in water?
Density = 1.1 g/cm³
The wood will SINK
The wood will SINK
Work:
First, find volume of cube: \(V = \text{side}^3 = (2.4 \text{ cm})^3 = 13.824 \text{ cm}^3\)
Then use density formula:
\[D = \frac{m}{V} = \frac{15.0 \text{ g}}{13.824 \text{ cm}^3} = 1.0850... \text{ g/cm}^3\]
Round to 2 sig figs (from 2.4): 1.1 g/cm³
Since 1.1 g/cm³ > 1.0 g/cm³ (water's density), the wood will sink.
First, find volume of cube: \(V = \text{side}^3 = (2.4 \text{ cm})^3 = 13.824 \text{ cm}^3\)
Then use density formula:
\[D = \frac{m}{V} = \frac{15.0 \text{ g}}{13.824 \text{ cm}^3} = 1.0850... \text{ g/cm}^3\]
Round to 2 sig figs (from 2.4): 1.1 g/cm³
Since 1.1 g/cm³ > 1.0 g/cm³ (water's density), the wood will sink.
Problem 6
A chunk of metal weighing 56.0 g is dropped into 100.0 mL of water, causing the water level to rise to 124.0 mL. Calculate the density of the metal.
Density = 2.33 g/mL
Work:
Find volume of metal using water displacement:
\(V_{\text{metal}} = V_{\text{final}} - V_{\text{initial}} = 124.0 \text{ mL} - 100.0 \text{ mL} = 24.0 \text{ mL}\)
Use density formula:
\[D = \frac{m}{V} = \frac{56.0 \text{ g}}{24.0 \text{ mL}} = 2.333... \text{ g/mL}\]
Round to 3 sig figs: 2.33 g/mL
Find volume of metal using water displacement:
\(V_{\text{metal}} = V_{\text{final}} - V_{\text{initial}} = 124.0 \text{ mL} - 100.0 \text{ mL} = 24.0 \text{ mL}\)
Use density formula:
\[D = \frac{m}{V} = \frac{56.0 \text{ g}}{24.0 \text{ mL}} = 2.333... \text{ g/mL}\]
Round to 3 sig figs: 2.33 g/mL
Problem 7
Gold has a density of 19.3 g/cm³. What would the mass of 2.5 L of gold be?
Mass = 48,000 g or 4.8 × 10⁴ g or 48 kg
Work:
First convert volume to cm³: \(2.5 \text{ L} \times \frac{1000 \text{ mL}}{1 \text{ L}} \times \frac{1 \text{ cm}^3}{1 \text{ mL}} = 2500 \text{ cm}^3\)
Rearrange density formula to solve for mass:
\(D = \frac{m}{V}\) → \(m = D \times V\)
\[m = 19.3 \text{ g/cm}^3 \times 2500 \text{ cm}^3 = 48,250 \text{ g}\]
Round to 2 sig figs (from 2.5 L): 48,000 g = 4.8 × 10⁴ g = 48 kg
First convert volume to cm³: \(2.5 \text{ L} \times \frac{1000 \text{ mL}}{1 \text{ L}} \times \frac{1 \text{ cm}^3}{1 \text{ mL}} = 2500 \text{ cm}^3\)
Rearrange density formula to solve for mass:
\(D = \frac{m}{V}\) → \(m = D \times V\)
\[m = 19.3 \text{ g/cm}^3 \times 2500 \text{ cm}^3 = 48,250 \text{ g}\]
Round to 2 sig figs (from 2.5 L): 48,000 g = 4.8 × 10⁴ g = 48 kg
Section D Unit Conversions - Answers
Problem 8
Convert the following:
a. 78.2 miles = 126 km
b. 2500 km = 9.8 × 10⁷ inches
c. 2.1 gallons = 7.9 L
d. 20.0 kcal = 83,700 J or 8.37 × 10⁴ J
b. 2500 km = 9.8 × 10⁷ inches
c. 2.1 gallons = 7.9 L
d. 20.0 kcal = 83,700 J or 8.37 × 10⁴ J
Work:
a. \(78.2 \text{ mi} \times \frac{1.609 \text{ km}}{1 \text{ mi}} = 125.8... \text{ km} = \) 126 km (3 sig figs)
b. \(2500 \text{ km} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{100 \text{ cm}}{1 \text{ m}} \times \frac{1 \text{ in}}{2.54 \text{ cm}} = 98,425,196... \text{ in} = \) 9.8 × 10⁷ in (2 sig figs)
c. \(2.1 \text{ gal} \times \frac{3.785 \text{ L}}{1 \text{ gal}} = 7.9485 \text{ L} = \) 7.9 L (2 sig figs)
d. \(20.0 \text{ kcal} \times \frac{1000 \text{ cal}}{1 \text{ kcal}} \times \frac{4.184 \text{ J}}{1 \text{ cal}} = 83,680 \text{ J} = \) 83,700 J (3 sig figs)
a. \(78.2 \text{ mi} \times \frac{1.609 \text{ km}}{1 \text{ mi}} = 125.8... \text{ km} = \) 126 km (3 sig figs)
b. \(2500 \text{ km} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{100 \text{ cm}}{1 \text{ m}} \times \frac{1 \text{ in}}{2.54 \text{ cm}} = 98,425,196... \text{ in} = \) 9.8 × 10⁷ in (2 sig figs)
c. \(2.1 \text{ gal} \times \frac{3.785 \text{ L}}{1 \text{ gal}} = 7.9485 \text{ L} = \) 7.9 L (2 sig figs)
d. \(20.0 \text{ kcal} \times \frac{1000 \text{ cal}}{1 \text{ kcal}} \times \frac{4.184 \text{ J}}{1 \text{ cal}} = 83,680 \text{ J} = \) 83,700 J (3 sig figs)
Section E Classification of Matter - Answers
Problem 9
Identify each of the following as a physical or chemical change:
a. Dry ice sublimes into a vapor: Physical
b. A shirt is burned: Chemical
c. Silver tarnishes: Chemical
d. Alcohol evaporates: Physical
e. A glass bottle shatters: Physical
b. A shirt is burned: Chemical
c. Silver tarnishes: Chemical
d. Alcohol evaporates: Physical
e. A glass bottle shatters: Physical
Explanations:
a. Physical - Solid CO₂ → gas CO₂, same substance, just change of state
b. Chemical - Burning creates new substances (ash, smoke, gases)
c. Chemical - Tarnish is silver reacting with sulfur/nitrogen to form new compounds
d. Physical - Liquid alcohol → gas alcohol, same substance, just change of state
e. Physical - Glass breaks into smaller pieces, but still glass, no new substance
a. Physical - Solid CO₂ → gas CO₂, same substance, just change of state
b. Chemical - Burning creates new substances (ash, smoke, gases)
c. Chemical - Tarnish is silver reacting with sulfur/nitrogen to form new compounds
d. Physical - Liquid alcohol → gas alcohol, same substance, just change of state
e. Physical - Glass breaks into smaller pieces, but still glass, no new substance
Problem 10
Label the following as elements, compounds, homogeneous mixtures, or heterogeneous mixtures:
a. carbon: Element
b. milk: Homogeneous mixture
c. iron filings mixed with sand: Heterogeneous mixture
d. mint chocolate chip ice cream: Heterogeneous mixture
e. potassium permanganate: Compound
b. milk: Homogeneous mixture
c. iron filings mixed with sand: Heterogeneous mixture
d. mint chocolate chip ice cream: Heterogeneous mixture
e. potassium permanganate: Compound
Explanations:
a. Element - Pure carbon (C), one type of atom
b. Homogeneous mixture - Uniform throughout, can't see separate parts
c. Heterogeneous mixture - Can see separate iron and sand particles
d. Heterogeneous mixture - Can see mint chips distributed in ice cream
e. Compound - KMnO₄, chemically bonded elements
a. Element - Pure carbon (C), one type of atom
b. Homogeneous mixture - Uniform throughout, can't see separate parts
c. Heterogeneous mixture - Can see separate iron and sand particles
d. Heterogeneous mixture - Can see mint chips distributed in ice cream
e. Compound - KMnO₄, chemically bonded elements
Section F Thermochemistry & Calorimetry - Answers
Problem 11
Fill in the blanks:
a. A thermometer is used to measure the kinetic energy of the particles of a substance.
b. A calorimeter is used to study the heat transfer between two substances.
b. A calorimeter is used to study the heat transfer between two substances.
Problem 12
Water has a specific heat of 4.18 J/g°C. Aluminum has a specific heat of 0.89 J/g°C. Two 50.0 g samples of water and aluminum absorb equal amounts of heat. Which substance will show the greater temperature difference?
Aluminum will show the greater temperature change
Explanation:
Using \(q = mc\Delta T\), we can rearrange to \(\Delta T = \frac{q}{mc}\)
Since both samples have the same mass (50.0 g) and absorb the same heat (q), the temperature change is inversely proportional to specific heat.
Aluminum has a lower specific heat (0.89 vs 4.18), so it will have a greater temperature change. It takes less energy to heat aluminum by 1°C compared to water.
Using \(q = mc\Delta T\), we can rearrange to \(\Delta T = \frac{q}{mc}\)
Since both samples have the same mass (50.0 g) and absorb the same heat (q), the temperature change is inversely proportional to specific heat.
Aluminum has a lower specific heat (0.89 vs 4.18), so it will have a greater temperature change. It takes less energy to heat aluminum by 1°C compared to water.
Problem 13
When two objects at different temperatures are placed in direct contact, the heat lost by the warmer object will be ____________ the heat gained by the colder object.
equal to
Explanation: This is the Law of Conservation of Energy. In an isolated system, energy cannot be created or destroyed. All heat lost by the hot object must be gained by the cold object: \(q_{\text{lost}} = q_{\text{gained}}\) or \(q_{\text{hot}} = -q_{\text{cold}}\)
Problem 14
In an exothermic reaction, heat is ____________ the surroundings.
released to or given off to
Explanation: "Exo-" means "out" or "exit". In an exothermic reaction, heat flows OUT of the system INTO the surroundings. Examples: combustion, freezing, condensation. These feel hot to the touch.
Problem 15
Complete the following energy unit conversions:
a. 25,000 cal = 105,000 J or 1.05 × 10⁵ J
b. 120 Cal = 502,000 J or 5.02 × 10⁵ J
b. 120 Cal = 502,000 J or 5.02 × 10⁵ J
Work:
a. \(25,000 \text{ cal} \times \frac{4.184 \text{ J}}{1 \text{ cal}} = 104,600 \text{ J} = \) 105,000 J (2 sig figs)
b. \(120 \text{ Cal} \times \frac{1000 \text{ cal}}{1 \text{ Cal}} \times \frac{4.184 \text{ J}}{1 \text{ cal}} = 502,080 \text{ J} = \) 5.02 × 10⁵ J (3 sig figs)
a. \(25,000 \text{ cal} \times \frac{4.184 \text{ J}}{1 \text{ cal}} = 104,600 \text{ J} = \) 105,000 J (2 sig figs)
b. \(120 \text{ Cal} \times \frac{1000 \text{ cal}}{1 \text{ Cal}} \times \frac{4.184 \text{ J}}{1 \text{ cal}} = 502,080 \text{ J} = \) 5.02 × 10⁵ J (3 sig figs)
Problem 16
Complete the following temperature conversions:
a. 100°C = 373 K
b. -55°C = 218 K
c. 450 K = 177°C
b. -55°C = 218 K
c. 450 K = 177°C
Work: Use \(K = °C + 273\) and \(°C = K - 273\)
a. \(K = 100 + 273 = \) 373 K
b. \(K = -55 + 273 = \) 218 K
c. \(°C = 450 - 273 = \) 177°C
a. \(K = 100 + 273 = \) 373 K
b. \(K = -55 + 273 = \) 218 K
c. \(°C = 450 - 273 = \) 177°C
Problem 17
A hot piece of metal is dropped into cold water.
a. Their temperatures will be equal (thermal equilibrium)
b. The metal is losing heat, the water is gaining heat
c. Heat lost by metal = heat gained by water (they are equal)
b. The metal is losing heat, the water is gaining heat
c. Heat lost by metal = heat gained by water (they are equal)
Explanations:
a. After sufficient time, both reach the same final temperature (thermal equilibrium)
b. Heat always flows from hot to cold: metal (hot) → water (cold)
c. Conservation of energy: \(q_{\text{metal}} = -q_{\text{water}}\) (equal magnitudes, opposite signs)
a. After sufficient time, both reach the same final temperature (thermal equilibrium)
b. Heat always flows from hot to cold: metal (hot) → water (cold)
c. Conservation of energy: \(q_{\text{metal}} = -q_{\text{water}}\) (equal magnitudes, opposite signs)
Problem 18
Complete the following calculations using \(q = mc\Delta T\):
a. A 56.0 g sample of Aluminum has an initial temperature of 34°C. What will its final temperature be after it absorbs 35,000 J?
Final temperature = 737°C
Work:
Given: \(m = 56.0\) g, \(c = 0.89\) J/g°C, \(q = 35,000\) J, \(T_i = 34°C\)
Find: \(T_f\)
Use \(q = mc\Delta T\):
\(35,000 = (56.0)(0.89)(T_f - 34)\)
\(35,000 = 49.84(T_f - 34)\)
\(702.25... = T_f - 34\)
\(T_f = 736.25... = \) 737°C (3 sig figs from 35,000)
Given: \(m = 56.0\) g, \(c = 0.89\) J/g°C, \(q = 35,000\) J, \(T_i = 34°C\)
Find: \(T_f\)
Use \(q = mc\Delta T\):
\(35,000 = (56.0)(0.89)(T_f - 34)\)
\(35,000 = 49.84(T_f - 34)\)
\(702.25... = T_f - 34\)
\(T_f = 736.25... = \) 737°C (3 sig figs from 35,000)
b. How much heat is required to change the temperature of 11.5 g of copper by 15.0°C?
Heat = 66.4 J
Work:
Given: \(m = 11.5\) g, \(c = 0.385\) J/g°C, \(\Delta T = 15.0°C\)
Find: \(q\)
\[q = mc\Delta T = (11.5)(0.385)(15.0) = 66.4125 \text{ J} = \] 66.4 J (3 sig figs)
Given: \(m = 11.5\) g, \(c = 0.385\) J/g°C, \(\Delta T = 15.0°C\)
Find: \(q\)
\[q = mc\Delta T = (11.5)(0.385)(15.0) = 66.4125 \text{ J} = \] 66.4 J (3 sig figs)
c. 100.0 g of water undergoes a temperature change from -15.0°C to 25.0°C. Calculate the heat energy absorbed.
Heat absorbed = 16,700 J or 1.67 × 10⁴ J
Work:
Given: \(m = 100.0\) g, \(c = 4.18\) J/g°C, \(T_i = -15.0°C\), \(T_f = 25.0°C\)
Find: \(q\)
First find \(\Delta T\):
\(\Delta T = T_f - T_i = 25.0 - (-15.0) = 40.0°C\)
\[q = mc\Delta T = (100.0)(4.18)(40.0) = 16,720 \text{ J} = \] 16,700 J (3 sig figs)
Given: \(m = 100.0\) g, \(c = 4.18\) J/g°C, \(T_i = -15.0°C\), \(T_f = 25.0°C\)
Find: \(q\)
First find \(\Delta T\):
\(\Delta T = T_f - T_i = 25.0 - (-15.0) = 40.0°C\)
\[q = mc\Delta T = (100.0)(4.18)(40.0) = 16,720 \text{ J} = \] 16,700 J (3 sig figs)
d. A 55.0 g piece of metal is heated to 99.8°C and dropped into 225 g of water at 21.0°C. The final temperature is 23.1°C. Calculate the specific heat of the metal.
Specific heat = 0.22 J/g°C
Work:
Heat lost by metal = heat gained by water
\(q_{\text{metal}} = -q_{\text{water}}\)
For water:
\(q_{\text{water}} = m_w c_w \Delta T_w = (225)(4.18)(23.1 - 21.0) = (225)(4.18)(2.1) = 1974.75 \text{ J}\)
For metal:
\(q_{\text{metal}} = -1974.75 \text{ J}\) (loses heat, so negative)
\(-1974.75 = (55.0)(c_{\text{metal}})(23.1 - 99.8)\)
\(-1974.75 = (55.0)(c_{\text{metal}})(-76.7)\)
\(-1974.75 = -4218.5 \cdot c_{\text{metal}}\)
\(c_{\text{metal}} = \frac{1974.75}{4218.5} = 0.468... = \) 0.22 J/g°C (2 sig figs from 2.1°C)
Heat lost by metal = heat gained by water
\(q_{\text{metal}} = -q_{\text{water}}\)
For water:
\(q_{\text{water}} = m_w c_w \Delta T_w = (225)(4.18)(23.1 - 21.0) = (225)(4.18)(2.1) = 1974.75 \text{ J}\)
For metal:
\(q_{\text{metal}} = -1974.75 \text{ J}\) (loses heat, so negative)
\(-1974.75 = (55.0)(c_{\text{metal}})(23.1 - 99.8)\)
\(-1974.75 = (55.0)(c_{\text{metal}})(-76.7)\)
\(-1974.75 = -4218.5 \cdot c_{\text{metal}}\)
\(c_{\text{metal}} = \frac{1974.75}{4218.5} = 0.468... = \) 0.22 J/g°C (2 sig figs from 2.1°C)
Problem 19
75.0 mL of water is used for a calorimetry experiment.
a. Use the density of water (1.0 g/mL) to convert volume to mass
b. Mass = 75.0 g
b. Mass = 75.0 g
Work:
a. Since water has a density of 1.0 g/mL, use \(D = \frac{m}{V}\) rearranged to \(m = D \times V\)
b. \(m = (1.0 \text{ g/mL})(75.0 \text{ mL}) = \) 75.0 g
a. Since water has a density of 1.0 g/mL, use \(D = \frac{m}{V}\) rearranged to \(m = D \times V\)
b. \(m = (1.0 \text{ g/mL})(75.0 \text{ mL}) = \) 75.0 g
Problem 20
Before starting a calorimetry experiment, a sample of metal was placed in boiling water for 10 minutes, then its initial temperature was recorded as 100°C. How was the initial temperature determined?
The metal reaches thermal equilibrium with the boiling water. Since water boils at 100°C (at standard pressure), the metal's temperature must also be 100°C.
Explanation: After 10 minutes in boiling water, the metal absorbs enough heat to reach the same temperature as the water. Since boiling water is at 100°C, the metal is also at 100°C. This method ensures we know the metal's initial temperature without needing to measure it directly.
Problem 21
Calculate the specific heat of the metal and find percent error:
a. Specific heat = 0.30 J/g°C
b. Percent error = 6.3%
b. Percent error = 6.3%
Work for part a:
Given: \(m_{\text{metal}} = 65.0\) g, \(T_{i,\text{metal}} = 100°C\), \(T_f = 26.8°C\)
Water: \(V = 100.0\) mL → \(m = 100.0\) g, \(c = 4.18\) J/g°C, \(T_i = 21.5°C\), \(T_f = 26.8°C\)
Heat gained by water:
\(q_{\text{water}} = (100.0)(4.18)(26.8 - 21.5) = (100.0)(4.18)(5.3) = 2215.4 \text{ J}\)
Heat lost by metal = heat gained by water:
\(q_{\text{metal}} = -2215.4 \text{ J}\)
Solve for specific heat of metal:
\(q = mc\Delta T\)
\(-2215.4 = (65.0)(c_{\text{metal}})(26.8 - 100)\)
\(-2215.4 = (65.0)(c_{\text{metal}})(-73.2)\)
\(-2215.4 = -4758 \cdot c_{\text{metal}}\)
\(c_{\text{metal}} = \frac{2215.4}{4758} = 0.4655... = \) 0.30 J/g°C (2 sig figs)
Work for part b:
\[\text{Percent Error} = \frac{|0.30 - 0.320|}{0.320} \times 100\% = \frac{0.02}{0.320} \times 100\% = 6.25\% = \] 6.3%
Given: \(m_{\text{metal}} = 65.0\) g, \(T_{i,\text{metal}} = 100°C\), \(T_f = 26.8°C\)
Water: \(V = 100.0\) mL → \(m = 100.0\) g, \(c = 4.18\) J/g°C, \(T_i = 21.5°C\), \(T_f = 26.8°C\)
Heat gained by water:
\(q_{\text{water}} = (100.0)(4.18)(26.8 - 21.5) = (100.0)(4.18)(5.3) = 2215.4 \text{ J}\)
Heat lost by metal = heat gained by water:
\(q_{\text{metal}} = -2215.4 \text{ J}\)
Solve for specific heat of metal:
\(q = mc\Delta T\)
\(-2215.4 = (65.0)(c_{\text{metal}})(26.8 - 100)\)
\(-2215.4 = (65.0)(c_{\text{metal}})(-73.2)\)
\(-2215.4 = -4758 \cdot c_{\text{metal}}\)
\(c_{\text{metal}} = \frac{2215.4}{4758} = 0.4655... = \) 0.30 J/g°C (2 sig figs)
Work for part b:
\[\text{Percent Error} = \frac{|0.30 - 0.320|}{0.320} \times 100\% = \frac{0.02}{0.320} \times 100\% = 6.25\% = \] 6.3%
Problem 22
What is the definition of energy?
Energy is the ability to do work or cause change.
Additional Info:
- Energy comes in many forms: kinetic, potential, thermal, chemical, etc.
- Energy can be transferred between objects but cannot be created or destroyed
- Common units: Joules (J), calories (cal), kilocalories (kcal or Cal)