14.2 Temperature Change and Heat Capacity

Temperature Change and Heat Capacity

Temperature change and heat capacity describe how energy transfers between objects cause those objects to get hotter or cooler. These concepts are the foundation for solving any problem where heat flows into or out of a substance, from warming a pot of water to designing a car's cooling system.

Temperature Change and Energy Transfer

Heat transfer always flows from a hotter object to a cooler one until both reach the same temperature (thermal equilibrium). Think of a hot coffee mug sitting on a table: heat flows from the mug into the surrounding air until the coffee cools to room temperature.

The change in a system's internal energy equals the amount of heat transferred to or from it:

• Positive heat transfer (heat flows in) increases internal energy and raises temperature, like a pot of water on a hot stove.
• Negative heat transfer (heat flows out) decreases internal energy and lowers temperature, like an ice cube absorbing heat from your drink as it melts.

The relationship connecting all of this is:

$Q = mc\Delta T$

where $Q$ is heat transferred, $m$ is mass, $c$ is specific heat capacity, and $\Delta T$ is the temperature change. This equation is the backbone of calorimetry, which is the measurement of heat transfer during physical or chemical changes.

Heat Calculations with Specific Heat

Every variable in $Q = mc\Delta T$ has a specific unit you need to keep straight:

• $Q$ = heat transferred (Joules, J)
• $m$ = mass of the object (kg)
• $c$ = specific heat capacity of the material (J/kg·K)
• $\Delta T$ = change in temperature (K or °C; the size of one degree is the same for both scales, so either works here)

Specific heat capacity ($c$) tells you how much energy is needed to raise the temperature of 1 kg of a substance by 1 K. A large $c$ means the substance resists temperature change; a small $c$ means it heats up or cools down easily.

Solving a heat transfer problem

1. Identify the mass ($m$), the specific heat capacity ($c$) of the material, and the desired temperature change ($\Delta T$).
2. Plug those values into $Q = mc\Delta T$.
3. Solve for $Q$.

Example: How much heat is needed to warm 1 kg of water by 10°C?

$Q = (1 \text{ kg})(4{,}186 \text{ J/kg·K})(10 \text{ K}) = 41{,}860 \text{ J}$

That's about 41.9 kJ, which gives you a feel for how much energy water can absorb.

Specific Heat Comparisons Across Substances

Different materials absorb heat at very different rates, and that's entirely because of their specific heat capacities.

Water's specific heat is remarkably high. A swimming pool can absorb a huge amount of solar energy during the day without its temperature rising much. Metals, on the other hand, have much lower values, which is why a metal pan on a stove gets scorching hot in seconds while the water inside it is still lukewarm.

These differences matter for real-world design:

• High specific heat materials are great for storing thermal energy and keeping temperatures stable. Water is used in central heating systems and car radiators for exactly this reason.
• Low specific heat materials respond quickly to temperature changes, making them useful for transferring heat fast. Aluminum and copper heat sinks in electronics pull heat away from components efficiently.

Choosing the right material for a job often comes down to whether you want to store heat or move it quickly.

Latent Heat and Phase Changes

During a phase change, a substance absorbs or releases energy without its temperature changing. This energy is called latent heat, and it goes toward breaking or forming the bonds between molecules rather than speeding them up.

• Heat of fusion ($L_f$): the energy needed to melt a solid into a liquid (or released when a liquid freezes). For water, $L_f = 334{,}000 \text{ J/kg}$.
• Heat of vaporization ($L_v$): the energy needed to boil a liquid into a gas (or released when a gas condenses). For water, $L_v = 2{,}256{,}000 \text{ J/kg}$.

The equation for latent heat is:

$Q = mL$

where $L$ is the relevant latent heat value. Notice there's no $\Delta T$ in this equation because the temperature stays constant during the phase change. This is why a pot of boiling water stays at 100°C no matter how long you keep the burner on; all the extra energy goes into converting liquid water into steam.