The First Law of Thermodynamics: Calculating Heat ($q$), Work ($w$), $\Delta E$, and $\Delta H$

Thermodynamics is often considered one of the most intimidating topics in chemistry and physics. It deals with the fundamental currency of the universe: energy. Whether you are calculating the fuel efficiency of an engine, determining the energy content of food, or predicting if a chemical reaction will explode, you are using the First Law of Thermodynamics.

At its core, the First Law is simple: Energy cannot be created or destroyed, only transferred.

However, applying this law to chemical systems requires mastering specific variables: heat ($q$), work ($w$), internal energy ($\Delta E$), and enthalpy ($\Delta H$). In this guide, we will break down these concepts, clarify the tricky sign conventions, and walk through step-by-step calculations using real exam-style problems.

Defining the System: Heat ($q$), Work ($w$), and Internal Energy ($\Delta E$)

Before doing any math, you must define your "universe." In thermodynamics, the universe is split into two parts:

• The System: The specific part of the universe you are studying (e.g., the gas inside a piston, the chemicals in a beaker).
• The Surroundings: Everything else (e.g., the water bath around the beaker, the laboratory, the atmosphere).

The internal energy ($\Delta E$) of a system changes when energy flows in or out across the boundary between the system and its surroundings. This flow happens in two forms: heat ($q$) and work ($w$).

Mastering Sign Conventions

The most common mistake students make is getting the signs wrong. Chemistry follows a "selfish system" perspective—everything is defined relative to the system's gain or loss.

• Heat ($q$):
  • (+) Positive: Heat flows into the system (Endothermic). The surroundings get cooler.
  • (-) Negative: Heat flows out of the system (Exothermic). The surroundings get warmer.
• Work ($w$):
  • (+) Positive: Work is done on the system (e.g., a piston compresses a gas).
  • (-) Negative: Work is done by the system (e.g., a gas expands and pushes a piston).

Illustrative Example: The Piston and the Water Bath

Let's look at a scenario involving a cylinder submerged in a water bath.

Problem: A chemical reaction occurs in a cylinder submerged in an insulated water bath. The reaction absorbs 287 kJ of heat from the water bath. Simultaneously, a piston pushes down on the gas, doing 207 kJ of work on the system. Calculate the change in internal energy ($\Delta E$) of the system. Does the water bath temperature go up or down?

Solution:

1. Analyze Signs:
   • Heat flows into the system: $q = +287 \text{ kJ}$.
   • Work is done on the system: $w = +207 \text{ kJ}$.
2. Calculate $\Delta E$: $$ \Delta E = q + w = 287 \text{ kJ} + 207 \text{ kJ} = 494 \text{ kJ} $$
3. Analyze the Water Bath: The problem asks about the water bath (the surroundings). Since the system absorbed heat ($q$ is positive), that heat had to come from the bath. Therefore, the bath loses energy, and its temperature goes down.

Practice Calculation: Expansion Work

Consider a different process carried out at constant pressure where $q = 200 \text{ kJ}$ and $w = -85 \text{ kJ}$.

• The negative work ($w = -85 \text{ kJ}$) implies the system expanded (did work on the surroundings).
• The positive heat ($q = 200 \text{ kJ}$) implies the process is endothermic.
• Internal Energy Change: $$ \Delta E = 200 + (-85) = 115 \text{ kJ} $$

Constant Pressure Processes: Enthalpy ($\Delta H$) and Hess's Law

In the laboratory, most chemical reactions happen in open beakers, meaning they occur at constant atmospheric pressure. When pressure is constant ($P_{ext} = \text{constant}$), a special relationship arises that simplifies our calculations.

We define a new quantity called Enthalpy ($H$). The change in enthalpy ($\Delta H$) is defined as:

However, the most important takeaway for general chemistry students is the practical definition:

This explains why chemists use $\Delta H$ and $q$ almost interchangeably. If you calculate the heat released by a reaction in an open beaker, you have calculated $\Delta H$.

Example: Relating $\Delta H$ to Heat and Work

Problem: A gas is confined in a cylinder at constant atmospheric pressure. You add 0.470 kJ of heat to the gas. It expands and does 211 J of work on the surroundings. What is $\Delta H$?

Analysis:

• Heat added ($q_p$) = $+0.470 \text{ kJ}$.
• Work done by system ($w$) = $-211 \text{ J}$ (or $-0.211 \text{ kJ}$).

Because the pressure is constant, we don't need to worry about the work term to find Enthalpy. By definition, $\Delta H = q_p$.

Result: $\Delta H = +0.470 \text{ kJ}$. (Note: $\Delta E$ would be $0.470 - 0.211 = 0.259 \text{ kJ}$).

Hess's Law: Adding Reaction Enthalpies

Enthalpy is a state function, meaning the path you take doesn't matter, only the start and end points. This leads to Hess's Law: If a reaction is carried out in a series of steps, $\Delta H$ for the overall reaction is the sum of $\Delta H$ for the individual steps.

Practice Problem:

Given the following data:

1. $A + B \rightarrow C \quad \Delta H = -100 \text{ kJ}$
2. $C \rightarrow D \quad \Delta H = +50 \text{ kJ}$

Calculate $\Delta H$ for the reaction $A + B \rightarrow D$.

Solution:

Add the two reactions together. Species $C$ appears as a product in reaction (1) and a reactant in reaction (2), so it cancels out:

Calorimetry and Specific Heat: Calculating Heat Transfer

How do we measure $q$ in the lab? We use calorimetry. The fundamental principle of calorimetry is the conservation of energy:

To relate heat to temperature change, we use Specific Heat Capacity ($c$), which is the energy required to raise 1 gram of a substance by 1°C.

Case Study: Final Temperature of a Mixture

This is a classic exam problem involving thermal equilibrium.

Problem: A 1.50 kg copper pipe at 800.0°C is dropped into a bucket containing 5.00 kg of water at 20.0°C. Calculate the final temperature ($T_f$).
Data: $c_{Cu} = 0.386 \text{ J/g}^\circ\text{C}$, $c_{H_2O} = 4.184 \text{ J/g}^\circ\text{C}$.

Step-by-Step Solution:

1. Identify the energy flow: Copper loses heat; Water gains heat.

2. Convert units and plug in numbers:
Masses must be in grams to match the specific heat units.

• Copper: $1500 \text{ g}$, $800.0^\circ\text{C}$, $0.386 \text{ J/g}^\circ\text{C}$
• Water: $5000 \text{ g}$, $20.0^\circ\text{C}$, $4.184 \text{ J/g}^\circ\text{C}$

3. Simplify the constants:

4. Expand and Solve for $T_f$:

Group $T_f$ terms on one side and constants on the other:

The final mixture reaches a lukewarm 41.0°C. This demonstrates the high specific heat of water—even a red-hot copper pipe only raises the water temperature by about 21 degrees!

Energy Changes in Phase Transitions and Chemical Reactions

Calculating heat gets slightly more complex when phase changes (melting, boiling) are involved. During a phase change, temperature remains constant even though heat is being added or removed. The energy is used to break intermolecular forces rather than increase kinetic energy.

For these steps, we use the formula: $q = n \Delta H_{\text{phase}}$, where $n$ is moles and $\Delta H$ is the enthalpy of fusion or vaporization.

Advanced Problem: Multi-Step Heating Curve

Problem: How much heat is evolved (released) when converting 2.00 mol of steam at 134°C to ice at -46°C?

Strategy: Break this into 5 distinct steps.

1. Cool Steam (134°C $\rightarrow$ 100°C)
2. Condense Steam to Liquid (Phase change at 100°C)
3. Cool Liquid Water (100°C $\rightarrow$ 0°C)
4. Freeze Liquid to Ice (Phase change at 0°C)
5. Cool Ice (0°C $\rightarrow$ -46°C)

Data provided:
Mass of 2.00 mol water = 36.03 g.
$c_{steam} = 2.01$, $c_{water} = 4.184$, $c_{ice} = 2.09$ (J/g°C).
$\Delta H_{vap} = 40.7 \text{ kJ/mol}$, $\Delta H_{fus} = 6.01 \text{ kJ/mol}$.

Calculations:

• Step 1 (Steam cooling): $$ q_1 = mc\Delta T = (36.03)(2.01)(34) = 2,462 \text{ J} = 2.46 \text{ kJ} $$
• Step 2 (Condensation): $$ q_2 = n\Delta H_{vap} = (2.00 \text{ mol})(40.7 \text{ kJ/mol}) = 81.4 \text{ kJ} $$
• Step 3 (Water cooling): $$ q_3 = mc\Delta T = (36.03)(4.184)(100) = 15,075 \text{ J} = 15.08 \text{ kJ} $$
• Step 4 (Freezing): $$ q_4 = n\Delta H_{fus} = (2.00 \text{ mol})(6.01 \text{ kJ/mol}) = 12.02 \text{ kJ} $$
• Step 5 (Ice cooling): $$ q_5 = mc\Delta T = (36.03)(2.09)(46) = 3,463 \text{ J} = 3.46 \text{ kJ} $$

Total Heat Evolved: $$ q_{total} = 2.46 + 81.4 + 15.08 + 12.02 + 3.46 = 114.42 \text{ kJ} $$ Rounding to 3 significant figures: 114 kJ.

Key Takeaways for Thermodynamics

• First Law: $\Delta E = q + w$. Always double-check your signs ($+$ for energy entering the system, $-$ for energy leaving).
• Enthalpy ($\Delta H$): At constant pressure, $\Delta H = q$. Positive $\Delta H$ is endothermic; Negative $\Delta H$ is exothermic.
• Calorimetry: Set heat lost = heat gained. Watch your units (Joules vs kJ, grams vs kg).
• Phase Changes: Temperature is constant during melting and boiling. Use $\Delta H_{fus}$ or $\Delta H_{vap}$ for these steps, and $mc\Delta T$ for the heating/cooling steps.