Specific heat capacity

Heat and temperature are not the same thing. Heat ($Q$) is energy in transit because of a temperature difference; it depends on how much substance is present, so it is an extensive property. Temperature is a measure of the degree of hotness — how energetically particles are moving on average — and does not depend on the amount of substance, so it is an intensive property. A large iceberg at 0°C holds far more heat energy than a lit match, even though the match is at a much higher temperature; size and quantity matter for heat, not for temperature.

Heat capacity $C$ of a particular body (not a substance in general) is the heat needed to raise its temperature by 1 K:

where: $Q$ is heat supplied in joules, $\Delta T$ is the temperature rise in kelvin (or °C, since a difference is numerically the same in both scales), and $C$ is in J/K. Heat capacity depends on the mass and material of that specific object — a large kettle full of water has a bigger heat capacity than a small cup of the same water.

Specific heat capacity $c$ fixes the mass at exactly 1 kg, making it a property of the material rather than of any one object:

where: $m$ is mass in kg, $\Delta T$ is temperature change in K, and $c$ is in J/kg·K. Rearranging gives the single working formula used in almost every calorimetry numerical: the heat gained or lost equals mass times specific heat capacity times temperature change. Water's specific heat capacity is about $4200\ \text{J/kg}\,\cdot\,\text{K}$ — roughly ten times that of most metals (copper is about $390\ \text{J/kg}\,\cdot\,\text{K}$, and even other metals rarely exceed $900\ \text{J/kg}\,\cdot\,\text{K}$). A kilogram of water therefore absorbs about ten times as much heat as a kilogram of metal for the same temperature rise. This single fact explains why water is used as a coolant in radiators (it carries away large amounts of heat without itself getting dangerously hot) and why seas and large lakes moderate nearby land temperatures (the water body warms up and cools down slowly, smoothing out the swings on the coast compared to inland areas).

Principle of method of mixtures. When a hot body and a cold body are mixed inside an insulated container (a calorimeter), heat flows only from the hot body to the cold body until both reach one common final temperature $T$. If no heat escapes to the surroundings:

where: $m_1, c_1, T_1$ describe the hot body and $m_2, c_2, T_2$ the cold body, with $T_1 > T > T_2$. This single equation is the key to every "find the final temperature" or "find the unknown specific heat capacity" numerical in this topic.

A piece of copper of mass $200\ \text{g}$ at $100°\text{C}$ is dropped into $150\ \text{g}$ of water at $25°\text{C}$ in a calorimeter of negligible heat capacity. Find the common final temperature, and check the answer using the derived formula. (Specific heat capacity of copper $= 390\ \text{J/kg}\,\cdot\,\text{K}$; of water $= 4200\ \text{J/kg}\,\cdot\,\text{K}$.)

Step 1. Convert masses to kilograms: $m_1 = 0.200\ \text{kg}$ (copper), $m_2 = 0.150\ \text{kg}$ (water); $T_1 = 100°\text{C}$, $T_2 = 25°\text{C}$.

Step 2. Apply the principle of method of mixtures directly. $0.200 \times 390 \times (100-T) = 0.150 \times 4200 \times (T-25)$, which simplifies to $78(100-T) = 630(T-25)$, giving $7800 - 78T = 630T - 15750$.

Step 3. Solve: $23550 = 708T$, so $T = 33.3°\text{C}$ (to one decimal place).

Step 4. Cross-check with the derived formula from above. $T = \dfrac{m_1c_1T_1+m_2c_2T_2}{m_1c_1+m_2c_2} = \dfrac{(78)(100)+(630)(25)}{78+630} = \dfrac{7800+15750}{708} = \dfrac{23550}{708} = 33.3°\text{C}$. Both routes agree.

Answer: the copper and water settle at a common temperature of $\mathbf{33.3°\text{C}}$, much closer to water's starting temperature than to copper's — because water's heat capacity $m_2c_2 = 630\ \text{J/K}$ is more than eight times copper's $m_1c_1 = 78\ \text{J/K}$, so water "dominates" the mixture, exactly as the weighted-average formula predicts.

Competitive stretch

If the same $200\ \text{g}$ copper piece at $100°\text{C}$ were instead dropped into just $50\ \text{g}$ of water at $25°\text{C}$, copper's heat capacity ($78\ \text{J/K}$) and water's new heat capacity ($0.050 \times 4200 = 210\ \text{J/K}$) become much closer, so $T = \dfrac{78(100)+210(25)}{78+210} = \dfrac{7800+5250}{288} = \dfrac{13050}{288} \approx 45.3°\text{C}$ — a noticeably higher final temperature than before, since the smaller mass of water can no longer dominate as strongly.

• Confusing heat and temperature. Heat depends on mass (extensive); temperature does not (intensive). "Hotter" is not the same as "more heat energy."
• Unit mismatches. Mass must be in kilograms and $c$ in J/kg·K for $Q$ to come out in joules; mixing grams with J/kg·K is the most common arithmetic slip in this chapter.
• Getting the subtraction order wrong. In $m_1c_1(T_1-T) = m_2c_2(T-T_2)$, both brackets must come out positive: the hot body's term is (higher minus final), the cold body's is (final minus lower).
• Assuming equal masses heat up equally. The same heat $Q$ produces very different $\Delta T$ in substances with different specific heat capacities — the reason a metal spoon heats almost instantly while the tea around it barely warms.

Ask the tutor: Want to try a three-substance mixing problem, or see how specific heat capacity connects to latent heat when a substance changes state during a calorimetry experiment — just ask.