Specific heat capacity measures how hard a substance is to warm up

Definition

Specific heat capacity

The energy required per unit mass per unit temperature increase.

\frac{\Delta E}{m\Delta\theta}

where $ΔE\Delta E$ is the energy in $J\text{J}$ , $m$ is the mass in $kg\text{kg}$ and $Δθ\Delta\theta$ is the temperature change in $∘C^\circ\text{C}$ , giving $c$ in $∘C)\text{J}/(\text{kg}\ ^\circ\text{C})$ .

Read the definition as an instruction: take one kilogram of the substance, warm it by one degree, and count the joules you had to supply.
A large specific heat capacity means the substance is stubborn: it takes a lot of energy to warm it, and it gives out a lot as it cools.
A small specific heat capacity means the substance heats and cools quickly, which is why a metal spoon left in a pan becomes too hot to touch long before the handle of the pan does.
The value is a property of the material, so doubling the mass of aluminium does not change $c$ , it only doubles the energy needed.
Two values worth remembering as a sense check are $∘C)4200\ \text{J}/(\text{kg}\ ^\circ\text{C})$ for water and about $∘C)900\ \text{J}/(\text{kg}\ ^\circ\text{C})$ for aluminium.

Practical

Investigation: the specific heat capacity of a metal block

Aim: to measure the specific heat capacity of aluminium by supplying a known amount of energy to a block of known mass.
Apparatus: eye protection, an aluminium block drilled with two holes, a $V12\ \text{V}$ immersion heater and power supply, a thermometer, a balance, a stopwatch, a heat-proof mat, insulation and a little oil.
Method:
- Find the mass $m$ of the dry block on the balance, in $kg\text{kg}$ , before anything is connected.
- Stand the block on the heat-proof mat, wrap it in insulation, and put a drop of oil in each hole so the heater and thermometer make good thermal contact with the metal.
- Slide the immersion heater into one hole and the thermometer into the other, and record the starting temperature $θ1\theta_1$ .
- Switch the heater on and start the stopwatch at the same moment, running it at a known power $P$ for a measured time $t$ of a few minutes.
- Switch off, keep watching the thermometer, and record the highest steady temperature $θ2\theta_2$ it reaches, since the metal keeps warming for a while after the current stops.
Measure and calculate:
- Energy supplied is $ΔE=Pt\Delta E = Pt$ , with $P$ in $W\text{W}$ and $t$ in $s\text{s}$ .
- Temperature rise is $Δθ=θ2−θ1\Delta\theta = \theta_2 - \theta_1$ , and the specific heat capacity follows from $\Delta E / (m\Delta\theta)$ , quoted to $2$ significant figures.
Result: aluminium comes out at roughly $∘C)900\ \text{J}/(\text{kg}\ ^\circ\text{C})$ , usually a little on the high side.
Watch out:
- Energy leaks to the surroundings, so the measured value comes out too high; lagging the block reduces the loss.
- Without the oil, the air gap between thermometer and metal is a poor conductor and the reading lags behind the true temperature.
- Some of the energy warms the heater itself and the thermometer, which is never counted in $P t$ .
Safety: wear eye protection, and never switch the heater on out of the block, because a bare element gets hot enough to burn within seconds.

Water is unusually stubborn, and that shapes the world around us

Water's specific heat capacity of $∘C)4200\ \text{J}/(\text{kg}\ ^\circ\text{C})$ is one of the largest of any everyday substance, roughly four to five times that of a typical metal.
That makes water an excellent coolant, because it can carry a lot of energy away from a car engine while its own temperature rises only modestly.
It makes water a good store in a central heating system, since a hot water tank holds a great deal of energy for its size.
It is also why the sea moderates the climate of an island like Britain: the water warms and cools far more slowly than the land, so coastal winters are milder and coastal summers cooler than inland.

Practical

Investigation: the specific heat capacity of water

Aim: to measure the specific heat capacity of water by heating a known mass of it electrically.
Apparatus: eye protection, a calorimeter or insulated beaker of water, a $V12\ \text{V}$ immersion heater and power supply, a thermometer, a balance, a stopwatch, a lid and lagging.
Method:
- Weigh the empty container, fill it with water, weigh it again, and subtract to get the mass of water $m$ in $kg\text{kg}$ .
- Lag the container and fit the lid, then lower the heater in so the element is fully covered by water and take the starting temperature $θ1\theta_1$ .
- Switch on and start the stopwatch together, heating at a known power $P$ for a measured time $t$ , stirring gently throughout so the water warms evenly.
- Switch off, keep stirring, and record the highest steady temperature $θ2\theta_2$ .
Measure and calculate: record $m$ , $P$ , $t$ , $θ1\theta_1$ and $θ2\theta_2$ , then use $ΔE=Pt\Delta E = Pt$ and $\Delta E / (m\Delta\theta)$ ; no graph is needed.
Result: water comes out at roughly $∘C)4200\ \text{J}/(\text{kg}\ ^\circ\text{C})$ , again a little high.
Watch out:
- Lag the container and use a lid, because energy lost to the room and by evaporation from the surface is the largest source of error.
- Stir before every reading, or the thermometer measures a hot layer near the heater rather than the whole body of water.
- Some energy warms the container itself, which the simple calculation ignores.
Safety: wear eye protection and keep the element under water at all times, and treat the hot water and hot heater as burn hazards.

Using the equation in calculations

The most useful rearrangement is $ΔE=mcΔθ\Delta E = mc\Delta\theta$ , which answers "how much energy is needed".
For the temperature change, use $Δθ=ΔE/(mc)\Delta\theta = \Delta E / (mc)$ , and for the mass, $\Delta E / (c\Delta\theta)$ .
Convert the mass into kilograms before substituting, because a mass in grams is out by a factor of a thousand.
$Δθ\Delta\theta$ is a difference, so it is the same number in $∘C^\circ\text{C}$ and in $K\text{K}$ and it never needs converting.
When a heater supplies the energy, get $ΔE\Delta E$ from $ΔE=Pt\Delta E = Pt$ with the time in seconds, which is where a question that quotes minutes tries to catch you out.

Common Mistake

Do not substitute a mass in grams, because $c$ is quoted per kilogram and a gram value gives an answer a thousand times too small.
Do not put the final temperature into $Δθ\Delta\theta$ , because the equation needs the temperature change.
Do not describe a measured value that is too high as "experimental error" and stop there; name energy lost to the surroundings as the reason.
Do not confuse specific heat capacity with the total energy an object stores, because $c$ belongs to the material and does not depend on the size of the sample.

Exam technique

You have to recall the equation, so write $\Delta E / (m\Delta\theta)$ first, then substitute, then evaluate, because a slip in the arithmetic still leaves the earlier marks intact.
Describing the experiment is point-marked, so name the balance for the mass, the power and the timed interval, both temperatures, and the insulation.
If asked for an improvement, give one that actually attacks the error you named, such as thicker lagging or a lid, rather than "repeat the experiment".
Give the unit in full as $∘C)\text{J}/(\text{kg}\ ^\circ\text{C})$ , since a bare number scores nothing on a "calculate the specific heat capacity" question.

Self review

Define specific heat capacity and give its unit.
How much energy is needed to raise the temperature of $kg2.0\ \text{kg}$ of water by $∘C30\ ^\circ\text{C}$ ?
In the metal block experiment, why is oil put into the thermometer hole?
Why does a measured specific heat capacity usually come out higher than the true value?
Give one everyday use of water that depends on its large specific heat capacity.

Specific heat capacity measures how hard a substance is to warm up

Definition

Specific heat capacity

The energy required per unit mass per unit temperature increase.

\frac{\Delta E}{m\Delta\theta}

Read the definition as an instruction: take one kilogram of the substance, warm it by one degree, and count the joules you had to supply.
A large specific heat capacity means the substance is stubborn: it takes a lot of energy to warm it, and it gives out a lot as it cools.
A small specific heat capacity means the substance heats and cools quickly, which is why a metal spoon left in a pan becomes too hot to touch long before the handle of the pan does.
The value is a property of the material, so doubling the mass of aluminium does not change $c$ , it only doubles the energy needed.
Two values worth remembering as a sense check are $∘C)4200\ \text{J}/(\text{kg}\ ^\circ\text{C})$ for water and about $∘C)900\ \text{J}/(\text{kg}\ ^\circ\text{C})$ for aluminium.

Practical

Investigation: the specific heat capacity of a metal block

Aim: to measure the specific heat capacity of aluminium by supplying a known amount of energy to a block of known mass.
Apparatus: eye protection, an aluminium block drilled with two holes, a $V12\ \text{V}$ immersion heater and power supply, a thermometer, a balance, a stopwatch, a heat-proof mat, insulation and a little oil.
Method:
- Find the mass $m$ of the dry block on the balance, in $kg\text{kg}$ , before anything is connected.
- Stand the block on the heat-proof mat, wrap it in insulation, and put a drop of oil in each hole so the heater and thermometer make good thermal contact with the metal.
- Slide the immersion heater into one hole and the thermometer into the other, and record the starting temperature $θ1\theta_1$ .
- Switch the heater on and start the stopwatch at the same moment, running it at a known power $P$ for a measured time $t$ of a few minutes.
- Switch off, keep watching the thermometer, and record the highest steady temperature $θ2\theta_2$ it reaches, since the metal keeps warming for a while after the current stops.
Measure and calculate:
- Energy supplied is $ΔE=Pt\Delta E = Pt$ , with $P$ in $W\text{W}$ and $t$ in $s\text{s}$ .
- Temperature rise is $Δθ=θ2−θ1\Delta\theta = \theta_2 - \theta_1$ , and the specific heat capacity follows from $\Delta E / (m\Delta\theta)$ , quoted to $2$ significant figures.
Result: aluminium comes out at roughly $∘C)900\ \text{J}/(\text{kg}\ ^\circ\text{C})$ , usually a little on the high side.
Watch out:
- Energy leaks to the surroundings, so the measured value comes out too high; lagging the block reduces the loss.
- Without the oil, the air gap between thermometer and metal is a poor conductor and the reading lags behind the true temperature.
- Some of the energy warms the heater itself and the thermometer, which is never counted in $P t$ .
Safety: wear eye protection, and never switch the heater on out of the block, because a bare element gets hot enough to burn within seconds.

Water is unusually stubborn, and that shapes the world around us

Water's specific heat capacity of $∘C)4200\ \text{J}/(\text{kg}\ ^\circ\text{C})$ is one of the largest of any everyday substance, roughly four to five times that of a typical metal.
That makes water an excellent coolant, because it can carry a lot of energy away from a car engine while its own temperature rises only modestly.
It makes water a good store in a central heating system, since a hot water tank holds a great deal of energy for its size.
It is also why the sea moderates the climate of an island like Britain: the water warms and cools far more slowly than the land, so coastal winters are milder and coastal summers cooler than inland.

Practical

Investigation: the specific heat capacity of water

Aim: to measure the specific heat capacity of water by heating a known mass of it electrically.
Apparatus: eye protection, a calorimeter or insulated beaker of water, a $V12\ \text{V}$ immersion heater and power supply, a thermometer, a balance, a stopwatch, a lid and lagging.
Method:
- Weigh the empty container, fill it with water, weigh it again, and subtract to get the mass of water $m$ in $kg\text{kg}$ .
- Lag the container and fit the lid, then lower the heater in so the element is fully covered by water and take the starting temperature $θ1\theta_1$ .
- Switch on and start the stopwatch together, heating at a known power $P$ for a measured time $t$ , stirring gently throughout so the water warms evenly.
- Switch off, keep stirring, and record the highest steady temperature $θ2\theta_2$ .
Measure and calculate: record $m$ , $P$ , $t$ , $θ1\theta_1$ and $θ2\theta_2$ , then use $ΔE=Pt\Delta E = Pt$ and $\Delta E / (m\Delta\theta)$ ; no graph is needed.
Result: water comes out at roughly $∘C)4200\ \text{J}/(\text{kg}\ ^\circ\text{C})$ , again a little high.
Watch out:
- Lag the container and use a lid, because energy lost to the room and by evaporation from the surface is the largest source of error.
- Stir before every reading, or the thermometer measures a hot layer near the heater rather than the whole body of water.
- Some energy warms the container itself, which the simple calculation ignores.
Safety: wear eye protection and keep the element under water at all times, and treat the hot water and hot heater as burn hazards.

Using the equation in calculations

The most useful rearrangement is $ΔE=mcΔθ\Delta E = mc\Delta\theta$ , which answers "how much energy is needed".
For the temperature change, use $Δθ=ΔE/(mc)\Delta\theta = \Delta E / (mc)$ , and for the mass, $\Delta E / (c\Delta\theta)$ .
Convert the mass into kilograms before substituting, because a mass in grams is out by a factor of a thousand.
$Δθ\Delta\theta$ is a difference, so it is the same number in $∘C^\circ\text{C}$ and in $K\text{K}$ and it never needs converting.
When a heater supplies the energy, get $ΔE\Delta E$ from $ΔE=Pt\Delta E = Pt$ with the time in seconds, which is where a question that quotes minutes tries to catch you out.

Common Mistake

Do not substitute a mass in grams, because $c$ is quoted per kilogram and a gram value gives an answer a thousand times too small.
Do not put the final temperature into $Δθ\Delta\theta$ , because the equation needs the temperature change.
Do not describe a measured value that is too high as "experimental error" and stop there; name energy lost to the surroundings as the reason.
Do not confuse specific heat capacity with the total energy an object stores, because $c$ belongs to the material and does not depend on the size of the sample.

Exam technique

You have to recall the equation, so write $\Delta E / (m\Delta\theta)$ first, then substitute, then evaluate, because a slip in the arithmetic still leaves the earlier marks intact.
Describing the experiment is point-marked, so name the balance for the mass, the power and the timed interval, both temperatures, and the insulation.
If asked for an improvement, give one that actually attacks the error you named, such as thicker lagging or a lid, rather than "repeat the experiment".
Give the unit in full as $∘C)\text{J}/(\text{kg}\ ^\circ\text{C})$ , since a bare number scores nothing on a "calculate the specific heat capacity" question.

Self review

Define specific heat capacity and give its unit.
How much energy is needed to raise the temperature of $kg2.0\ \text{kg}$ of water by $∘C30\ ^\circ\text{C}$ ?
In the metal block experiment, why is oil put into the thermometer hole?
Why does a measured specific heat capacity usually come out higher than the true value?
Give one everyday use of water that depends on its large specific heat capacity.

2.2.2b Specific heat capacity

Specific heat capacity measures how hard a substance is to warm up

Investigation: the specific heat capacity of a metal block

Water is unusually stubborn, and that shapes the world around us

Investigation: the specific heat capacity of water

Using the equation in calculations

Review 2.2.2b Specific heat capacity by teaching Genie

2.2.2b Specific heat capacity

Specific heat capacity measures how hard a substance is to warm up

Investigation: the specific heat capacity of a metal block

Water is unusually stubborn, and that shapes the world around us

Investigation: the specific heat capacity of water

Using the equation in calculations

Review 2.2.2b Specific heat capacity by teaching Genie

2.2.2b Specific heat capacity

Specific heat capacity measures how hard a substance is to warm up

Investigation: the specific heat capacity of a metal block

Water is unusually stubborn, and that shapes the world around us

Investigation: the specific heat capacity of water

Using the equation in calculations

Review 2.2.2b Specific heat capacity by teaching Genie

2.2.2b Specific heat capacity Revision Guide

2.2.2b Specific heat capacity

Specific heat capacity measures how hard a substance is to warm up

Investigation: the specific heat capacity of a metal block

Water is unusually stubborn, and that shapes the world around us

Investigation: the specific heat capacity of water

Using the equation in calculations

Review 2.2.2b Specific heat capacity by teaching Genie

2.2.2b Specific heat capacity Revision Guide