Specific heat capacity and thermal balance

Heat transfer is a change of internal energy without doing work: a body receives or gives up energy, and this shows up as a change in its temperature. The specific heat capacity $c$ tells how much heat must be transferred to $1\ \text{kg}$ of a substance to change its temperature by $1\ \text{K}$:

$$Q = cm\,\Delta T = cm\,(t_2 - t_1).$$

When heating, $t_2 > t_1$ and $Q > 0$; when cooling, $Q < 0$. The product $C = cm$ is called the heat capacity of the body: it grows with mass, whereas $c$ depends only on the substance and its state. If the bodies exchange heat only with each other, the total internal energy of the system is conserved, and the heat-balance equation reads

$$Q_1 + Q_2 + \dots = 0 \quad\text{or}\quad Q_{\text{lost}} = Q_{\text{gained}}.$$

From this equation one finds the final temperature of a mixture, an unknown mass, or a specific heat capacity.

An aluminium part of mass $m = 0{,}50\ \text{kg}$ is heated from $t_1 = 20\ ^\circ\text{C}$ to $t_2 = 120\ ^\circ\text{C}$. The specific heat capacity of aluminium is $c = 900\ \text{J}/(\text{kg}\cdot\text{K})$. How much heat must be supplied to the part?

In an insulated vessel, $m_1 = 0{,}30\ \text{kg}$ of water at $t_1 = 90\ ^\circ\text{C}$ is mixed with $m_2 = 0{,}20\ \text{kg}$ of water at $t_2 = 20\ ^\circ\text{C}$. The specific heat capacity of water is $c = 4200\ \text{J}/(\text{kg}\cdot\text{K})$. Find the equilibrium temperature $t$.

An aluminium block of mass $m_a = 0{,}15\ \text{kg}$, heated to $t_a = 200\ ^\circ\text{C}$, is placed into a calorimeter with water of mass $m_w = 0{,}40\ \text{kg}$ at $t_w = 15\ ^\circ\text{C}$. Heat losses and the heat capacity of the calorimeter are neglected. Given $c_a = 900\ \text{J}/(\text{kg}\cdot\text{K})$ and $c_w = 4200\ \text{J}/(\text{kg}\cdot\text{K})$. Find the final temperature $t$ and the heat transferred from the block to the water.

Self-study plan (model first; a real experiment is optional).

1. Open the PhET simulation “Energy Forms and Changes” or any online calorimeter model.
2. Set two samples of the same substance with different initial temperatures and masses, start the process, and record the thermometer readings before and after equilibrium.
3. Put these numbers into a table in your notebook or a spreadsheet and compute $Q_{\text{lost}}$ and $Q_{\text{gained}}$ with $Q = cm\,\Delta T$. Verify that $Q_{\text{lost}} \approx Q_{\text{gained}}$.
4. Change the mass of one body by a factor of 2, predict the new final temperature with $m_1(t_1 - t) = m_2(t - t_2)$, and then check the prediction in the model.
5. If you want a real experiment, use only warm water no hotter than $40\ ^\circ\text{C}$ in plastic cups: no boiling water, no open flame, no glassware near the table edge; wipe up spilled water at once.

To show that in an insulated system heat flows from the hot body to the cold one, that the heat lost equals the heat gained, and that the final temperature settles between the initial temperatures.

Diagram + graph. On the left, a calorimeter containing a hot body (red) and a cold body (blue); the arrow $Q_{\text{lost}}$ points from the hot body to the cold one, the arrow $Q_{\text{gained}}$ points to the cold body, with the label $Q_{\text{lost}} = Q_{\text{gained}}$ beside them. On the right, a graph of temperature versus time $t(\tau)$: the curve of the hot body falls, the curve of the cold body rises, and both approach the horizontal line $t = t_{\text{eq}}$, where $t_2 < t_{\text{eq}} < t_1$.

A kettle is filled with $1{.}5$ L of water at $20$ °C and switched on with a $2000$ W heater. The specific heat capacity of water is $c=4200$ J/(kg·°C), its density is $\rho=1000$ kg/m³, so the water mass is $m=\rho V=1{.}5$ kg. To heat it to $100$ °C requires $\Delta T=80$ °C.

This is a direct use of $Q=cm\Delta T$: we find the heat the water receives from the heater. With $P=2000$ W the heating time is $t=Q/P$: $Q=4200\cdot1{.}5\cdot80=504{,}000$ J, so $t\approx504{,}000/2000=252$ s.

An athlete drinks $0{.}5$ kg of water at $10$ °C. The body warms it to $37$ °C. The specific heat capacity of water is $c=4200$ J/(kg·°C).

Here $Q=cm\Delta T$ applies with $\Delta T=27$ °C: $Q=4200\cdot0{.}5\cdot27=56{,}700$ J. This heat is removed from the body, which helps prevent overheating.

A car engine cooling system circulates $5$ kg of antifreeze with specific heat capacity $c\approx3800$ J/(kg·°C). During a working cycle it heats from $20$ °C to $95$ °C.

The heat carried away from the engine by the liquid is $Q=cm\Delta T=3800\cdot5\cdot75=1{,}425{,}000$ J $=1{.}425$ MJ. A large specific heat capacity lets it remove more heat for a moderate temperature rise.