A pack of crackers lists its energy in food Calories (Cal), an electric kettle is rated in kW, and an instant cold pack turns cold in your hand. All three are about energy: how much there is, how fast it is delivered, and which way it flows.

This lesson builds the language for the whole thermochemistry unit. Any energy can be written in J, converted to another unit with an exact factor, and given a sign that says whether it enters or leaves the system (Figure 1).

1. Energy in any unitcal, kcal (Cal), erg, N·m
2. $\times$unit factor: into J
3. Joules (J)the SI unit of energy
4. $\times$unit factor: into the unit asked for
5. Wanted unitwith a sign: + in, − out of the system

Given a power and a time? Find the energy first: $E = P \times t$.Exact factors: $1\,\text{cal} = 4.184\,\text{J}$, $1\,\text{erg} = 10^{-7}\,\text{J}$.

Learning Objectives

After this lesson, you should be able to:

1. Convert between energy units (J, cal, erg, N m) and tell energy from power.
2. Use system and surroundings and sign conventions; classify processes as exothermic or endothermic.

Key Ideas

1. Energy, Work, and Heat

Energy is the capacity to do work or to transfer heat. Work is done when a force moves an object; for a constant force pushing along the direction of motion, the work is the force times the distance. Heat is energy that moves because of a temperature difference, always from the hotter object to the colder one.

• $F$ = force, in N
• $d$ = distance moved along the force, in m
• $w$ = work, in J

1 N × 1 m = 1 J, and the joule (J) is the SI unit of energy. Pushing a crate with a force of 25.0 N across 3.20 m transfers 80.0 J of energy as work (Figure 2, Panel (a)).

Key idea: work and heat are the two ways energy moves between objects, and both are measured in the same unit, the joule.

2. Units of Energy: Joule, Calorie, Erg

Chemistry uses several energy units. The calorie (cal) is defined as exactly 4.184 J. The Calorie with a capital C on food labels is a kilocalorie. The erg is the much smaller energy unit of the older centimeter-gram-second (CGS) system, still found in some physics tables.

On a number line the units span many powers of ten (Figure 3). To convert, write the equality as a fraction (a unit factor) and multiply, placing the unit you start with in the denominator so that it cancels.

Exam tip: the factors 4.184, 1000 and 10⁷ are exact definitions, so they never limit the number of significant figures in an answer.

3. Energy versus Power

Power is the rate at which energy is transferred: energy per unit time. Its SI unit is the watt (W): 1 W = 1 J/s. A watt is not an energy unit; multiply a power by a time to get an energy.

In one minute a 1.50 kW kettle delivers 90.0 kJ, while a 0.100 kW electric fan uses only 6.00 kJ (Figure 4): the bigger the power, the steeper the energy line. The kilowatt-hour on an electricity bill is an energy unit for the same reason: $1\,\text{kW·h} = 1000\,\text{J/s} \times 3600\,\text{s} = 3.6 \times 10^{6}\,\text{J}$.

Exam tip: a common test item asks which unit is not an energy unit. Joule, calorie, erg, newton meter and kilowatt-hour are all energy units; the watt is a unit of power.

4. System, Surroundings, and Signs

The system is the part of the universe we choose to study, such as the chemicals in a beaker or the gas in a cylinder. The surroundings are everything else. Energy is never created or destroyed; it only crosses the boundary between the two, and its sign tells the direction, always from the point of view of the system (Figure 5: Panel (a) energy in, Panel (b) energy out).

• $q > 0$: heat is absorbed by the system; the process is endothermic.
• $q < 0$: heat is released by the system; the process is exothermic.
• $w > 0$: work is done on the system by the surroundings.
• $w < 0$: work is done by the system on the surroundings.

You can tell the two kinds of process apart by touch: an exothermic process warms its surroundings, such as burning fuel or an activated hand warmer (Figure 6, Panel (a)), and an endothermic process cools them, such as melting ice or an instant cold pack (Panel (b)).

Key idea: the sign always describes the system. Energy in is positive; energy out is negative.

Worked Example 1: Food Calories (Cal) to kJ

Problem: A serving of crackers lists 185 Cal (food Calories) on its nutrition label. What is this energy in kJ?

Step 1. Write the unit equality. A food Calorie is a kilocalorie, and 1 kcal is 4.184 kJ.

Step 2. Multiply by the unit factor that cancels Cal.

Final Answer: 774 kJ

The answer keeps three significant figures, set by 185 Cal; the factor 4.184 is exact.

1. Energy given$185\,\text{Cal}$
2. $\times \dfrac{4.184\,\text{kJ}}{1\,\text{Cal}}$Step 2: Cal cancels
3. Energy$774.0\,\text{kJ} \approx 774\,\text{kJ}$Final answer to 3 significant figures, set by 185 Cal

Your turn: convert between energy units

Same two steps as Worked Example 1: write the unit equality, then multiply by the factor that cancels the starting unit.

Worked Example 2: Ergs to Calories

Problem: An old physics handbook that uses CGS units gives the kinetic energy of a served tennis ball as 6.85 × 10⁸ erg. Express this energy in cal.

Step 1. Convert erg to J. 1 erg is $10^{-7}$ J.

Step 2. Convert J to cal.

Final Answer: 16.4 cal

Going through J keeps every factor simple: one power-of-ten shift, then the familiar 4.184.

1. Energy given$6.85 \times 10^{8}\,\text{erg}$
2. $\times \dfrac{10^{-7}\,\text{J}}{1\,\text{erg}}$Step 1: erg to J
3. Energy in J$68.5\,\text{J}$
4. $\times \dfrac{1\,\text{cal}}{4.184\,\text{J}}$Step 2: J to cal
5. Energy$16.37\,\text{cal} \approx 16.4\,\text{cal}$Final answer to 3 significant figures

Your turn: erg to cal

Same method as Worked Example 2: multiply the erg value by $10^{-7}$ to get J (1 erg = $10^{-7}$ J), then divide the J value by 4.184 J/cal.

Worked Example 3: Energy from Power and Time

Problem: An electric kettle rated at 1.85 kW is switched on for 2.40 min. How much electrical energy, in kJ, does it use?

Step 1. Write the power as energy per second.

Step 2. Convert the time to seconds.

Step 3. Multiply the power by the time.

Final Answer: 266 kJ

The s units cancel and leave kJ. In J this is 2.66 × 10⁵ J, which is why kJ is the convenient unit here.

1. Time$2.40\,\text{min}$
2. $\times \dfrac{60\,\text{s}}{1\,\text{min}}$Step 2: min to s
3. Time$144.0\,\text{s}$
4. $\times 1.85\,\text{kJ/s}$Step 3: power times time
5. Energy used$266.4\,\text{kJ} \approx 266\,\text{kJ}$Final answer to 3 significant figures

Your turn: energy from power and time

Same three steps as Worked Example 3: kW as kJ/s, minutes to seconds, then E = P × t.

Worked Example 4: Sign of q for an Exothermic Process

Problem: An iron-powder hand warmer (the system) gives off 6.35 kcal of heat to your hands. What is q for the hand warmer, in kJ?

Step 1. Decide the direction of the energy transfer. Heat leaves the hand warmer and flows into your hands, so q is negative: the process is exothermic.

Step 2. Convert the amount of heat to kJ.

Step 3. Attach the sign.

Final Answer: q = −26.6 kJ

Your hands, part of the surroundings, gain the same 26.6 kJ: for them the sign would be positive.

1. Heat given off$6.35\,\text{kcal}$
2. $\times \dfrac{4.184\,\text{kJ}}{1\,\text{kcal}}$Step 2: size in kJ
3. Size$\left| q \right| = 26.57\,\text{kJ}$
4. $\text{heat out, so } q \text{ is negative}$Step 3: heat leaves the system
5. q for the hand warmer$q = -26.57\,\text{kJ} \approx -26.6\,\text{kJ}$Exothermic; final answer to 3 significant figures

Your turn: sign of q or w

Same three steps as Worked Example 4: decide the direction, convert the amount, attach the sign. Type the minus sign when energy leaves the system.

Common Mistakes

Common mistake: Turning the unit factor upside down. In Worked Example 1, dividing 185 Cal by 4.184 gives 44.22, far too small. Put the unit you start with in the denominator so it cancels.

Common mistake: Mixing up Cal, cal and kcal. 1 Cal = 1 kcal = 1000 cal. Treating 185 Cal as 185 cal gives 0.774 kJ instead of 774 kJ. Also check the prefix of the unit asked for: kJ and J differ by 1000.

Common mistake: Getting the erg power of ten backwards. One erg is a tiny amount of energy, 10⁻⁷ J, so the value in J is much smaller than the value in erg. Multiplying by 10⁷ instead, or skipping the J step, gives an answer off by many powers of ten.

Common mistake: Treating watts as energy, or forgetting the seconds. 1 W = 1 J/s. In Worked Example 3, $1.85 \times 2.40 = 4.44$ uses minutes, not seconds; convert the time to seconds first, then multiply.

Common mistake: Reversing the sign. The sign belongs to the system. Heat released by the system is negative, even though the surroundings warm up.

Common mistake: Rounding too early. Keep 4 significant figures in intermediate values and round only the final answer to 3.

Key Takeaway

Energy can be measured in J, cal, kcal (food Calories, Cal), erg or N·m, and exact unit factors convert between them. Power is energy per time, so E = P × t. The sign of q or w tells whether energy enters (positive) or leaves (negative) the system: heat released means exothermic, heat absorbed means endothermic.

Practice: randomized problems

Each problem below uses fresh numbers every time and mixes every kind of problem in this lesson: work it out, check your answer, then open the worked solution to compare your steps.

Your results

This summary collects every practice problem you try on this page and suggests what to review next.

Figures: original diagrams by Dr. Ray’s Learning Center, licensed CC BY 4.0. Conversion factors and definitions: Whitten, Davis, Peck and Stanley, Chemistry, 9th ed. (Brooks/Cole, Cengage Learning, 2010). The practice problems are generated and checked by an audited engine; see the lesson audit log.