5.1.3 - Specific heat capacity and latent heat

Specific Heat Capacity

Different materials need different amounts of energy for the same temperature rise. A pan made from a low heat capacity metal warms quickly; the same mass of water warms more slowly because water has a much larger specific heat capacity.

The OCR thermal-physics relationship is:

The interval is what matters: a rise of $15\,\mathrm{K}$ is the same size as a rise of $15\,{}^\circ\mathrm C$.

Read the equation as a statement about proportionality. For the same material, doubling the mass doubles the energy needed. For the same mass, doubling the temperature rise doubles the energy needed. For the same mass and temperature rise, the material with larger $c$ needs more energy.

Worked example: heating a metal block

A $0.42\,\mathrm{kg}$ aluminium block is heated from $19\,^{\circ}\mathrm{C}$ to $37\,^{\circ}\mathrm{C}$. Take the specific heat capacity of aluminium as $910\,\mathrm{J}\,\mathrm{kg}^{-1}\,\mathrm{K}^{-1}$. Calculate the energy transferred to the block.

The temperature change is:

Then:

To two significant figures, the energy transferred is:

Notice that the calculation uses a temperature change, not the final temperature itself.

Electrical Method For Specific Heat Capacity

An electrical method is useful because the energy supplied by a heater can be calculated from electrical measurements. If the p.d. across the heater is $V$, the current through it is $I$, and it is switched on for time $t$, then:

If the energy all went into the sample:

For a metal block:

1. Measure the mass of the block.
2. Place a low-voltage immersion heater and a temperature probe or thermometer into holes in the block.
3. Add a small amount of oil or water in the holes to improve thermal contact, if appropriate for the apparatus.
4. Wrap the block in insulation and record the initial temperature.
5. Switch on the heater, measure $V$, $I$, and $t$, then record the final temperature.
6. Use $c = VIt/(m\Delta\theta)$.

For a liquid, the idea is the same, but the liquid should be in an insulated container with a lid. Measure the mass of the liquid, keep the heater fully immersed, and stir gently so the temperature probe measures a representative temperature rather than a hot region near the heater.

There is also a mixture method for estimating specific heat capacity. A hot object of known mass and temperature is transferred into a cooler liquid of known mass and temperature. If heat losses and the calorimeter's heat capacity are small or corrected for, then:

This method is an estimate because transferring the hot object takes time and some energy is lost to the surroundings. OCR's additional guidance names it because it is a good way to test energy conservation and experimental evaluation.

In the electrical method, a graph can improve the analysis. If $V$ and $I$ are constant, plotting temperature rise $\Delta\theta$ against time $t$ should give an approximately straight line. Since:

the gradient is $VI/(mc)$, so:

An alternative is to plot electrical energy $VIt$ on the vertical axis against temperature rise on the horizontal axis. The gradient is then $mc$, so $c = \text{gradient}/m$.

Using a mixture quantitatively. A $0.100\,\mathrm{kg}$ metal sample at $90.0\,{}^\circ\mathrm C$ is placed into $0.200\,\mathrm{kg}$ of water at $20.0\,{}^\circ\mathrm C$. The final temperature is $23.0\,{}^\circ\mathrm C$. Neglect heat loss and the container's heat capacity. With $c_w=4200\,\mathrm{J\,kg^{-1}\,K^{-1}}$,

The two temperature changes differ. If the container absorbs appreciable energy, add $C_{\text{container}}(23.0-20.0)$ to the right-hand side; otherwise the estimate of the metal's specific heat capacity is too low.

[DIAGRAM: asset_slug: 5_01_3_specific_heat_capacity_and_latent_heat__diagram_01; description: Electrical heating arrangements for a metal block and a liquid; measure current in series and potential difference across the heater.; caption: Electrical heating arrangements measure current through the heater and potential difference across it.]

Specific Latent Heat

When a substance changes state, energy is transferred even though the temperature can stay constant. Melting ice at $0\,^{\circ}\mathrm{C}$ produces water at $0\,^{\circ}\mathrm{C}$; boiling water at $100\,^{\circ}\mathrm{C}$ produces steam at $100\,^{\circ}\mathrm{C}$ at normal pressure. The energy is not raising temperature during that stage. It is changing the arrangement and separation of the molecules.

There are two OCR terms you must keep separate:

• specific latent heat of fusion: solid to liquid, or liquid to solid
• specific latent heat of vaporisation: liquid to gas, or gas to liquid

The equation is:

Worked example: melting ice at its melting point

A $0.085\,\mathrm{kg}$ sample of ice at $0\,^{\circ}\mathrm{C}$ melts completely to water at $0\,^{\circ}\mathrm{C}$. Take the specific latent heat of fusion of ice as $3.34\times 10^{5}\,\mathrm{J}\,\mathrm{kg}^{-1}$. Calculate the energy required.

Because the temperature does not change, use $E = mL$, not $mc\Delta\theta$.

To three significant figures: