Heat Calculations (q = mcΔT) - Chemistry

1. Fundamental Concepts

Heat (): The energy transferred between a system and its surroundings due to a temperature difference. The SI unit is the joule (J), and the calorie (cal) is also commonly used, with the conversion factor . Heat flows spontaneously from an object at a higher temperature to one at a lower temperature.
Mass (): The mass of the substance involved in the heat calculation. The standard unit is grams (g), while kilograms (kg) may be used in advanced contexts. Units must be consistent throughout the calculation.
Specific Heat Capacity (): The amount of heat required to raise the temperature of one unit mass of a pure substance by one temperature unit. It is an intrinsic physical property of a substance. The common unit is , and its value is independent of the mass and temperature change of the sample.
Temperature Change (): Defined as the final temperature minus the initial temperature, expressed as , with units of degrees Celsius (). A positive indicates heat absorption (positive ), and a negative indicates heat release (negative ).

2. Key Concepts

Core Formula: The fundamental equation for heat calculations is . It applies only to pure substances undergoing temperature changes with no phase changes or chemical reactions. For phase changes such as melting, freezing, vaporization, and condensation, this equation is insufficient, and latent heat of phase change must be calculated separately.
Unit Consistency: All units for , , and must be compatible. For example, if specific heat is given in , mass must be in grams and temperature in degrees Celsius to yield heat in joules.
Heat Conservation: In an adiabatic system (no heat exchange with the surroundings), the heat released by a hotter substance equals the heat absorbed by a colder substance, expressed as . This principle is used for heat calculations involving mixed substances.
Substance-Specific Property: Different substances have distinct specific heat capacities. For instance, liquid water has a specific heat capacity of approximately , one of the highest among common substances, which explains water’s ability to moderate temperature.

3. Examples

Easy

Calculate the heat absorbed when 50.0 g of liquid water is heated from 25.0 $^\circ\text{C}$ to 50.0 $^\circ\text{C}$. The specific heat capacity of water is $c = 4.184\ \text{J}/(\text{g}\cdot^\circ\text{C})$.

1. Calculate the temperature change: $\Delta T = 50.0 - 25.0 = 25.0\ ^\circ\text{C}$

2. Substitute into the formula: $q = 50.0\ \text{g} \times 4.184\ \text{J}/(\text{g}\cdot^\circ\text{C}) \times 25.0\ ^\circ\text{C} = 5230\ \text{J}$

Medium

A 100.0 g sample of aluminum metal, initially at 100.0 $^\circ\text{C}$, cools to 25.0 $^\circ\text{C}$ in a room-temperature environment. The specific heat capacity of aluminum is $c = 0.900\ \text{J}/(\text{g}\cdot^\circ\text{C})$. Determine the heat released.

1. Calculate the temperature change: $\Delta T = 25.0 - 100.0 = -75.0\ ^\circ\text{C}$

2. Substitute into the formula: $q = 100.0\ \text{g} \times 0.900\ \text{J}/(\text{g}\cdot^\circ\text{C}) \times (-75.0)\ ^\circ\text{C} = -6750\ \text{J}$. The negative sign indicates that aluminum releases heat, with a magnitude of 6750 J.

Hard
80.0 g of hot water at 95.0 $^\circ\text{C}$ is mixed with 20.0 g of cold water at 10.0 $^\circ\text{C}$. Assuming no heat loss to the surroundings, find the final equilibrium temperature of the mixture. The specific heat capacity of water is $c = 4.184\ \text{J}/(\text{g}\cdot^\circ\text{C})$.

1. Let the final temperature be $T$. By heat conservation: $q_{\text{hot}} = -q_{\text{cold}}$

2. Substitute and expand: $80.0 \times 4.184 \times (T - 95.0) = -[20.0 \times 4.184 \times (T - 10.0)]$

3. Cancel the common specific heat term and solve for $T$: $T = 78.0\ ^\circ\text{C}$

4. Problem-Solving Techniques

Identify Known and Unknown Quantities: First, list all given values of $m$, $c$, $T_{\text{initial}}$, and $T_{\text{final}}$, and clearly define the unknown variable. Rearrange the core formula as needed:

$m = \dfrac{q}{c\Delta T}$, $c = \dfrac{q}{m\Delta T}$, $\Delta T = \dfrac{q}{mc}$.

Verify Unit Consistency: Standardize all units before performing calculations. Pay close attention to mass units (grams vs. kilograms) and temperature units to avoid errors from mismatched units.

Interpret the Sign of Heat: Determine whether the substance absorbs or releases heat before calculation. The sign of $\Delta T$ determines the sign of $q$, which can be used to verify the physical validity of the final result.

Approach for Mixture Problems: For adiabatic mixing processes, apply the principle of heat conservation first. Write separate heat expressions for each component, set up the equilibrium equation, and cancel common terms (such as specific heat capacity) to simplify calculations.

Check for Phase Changes: If a problem mentions melting, freezing, vaporization, or condensation, confirm whether the temperature reaches the phase transition point. If a phase change occurs, calculate the latent heat of the phase change separately, then combine it with the sensible heat calculated from $q = mc\Delta T$.