Determine specific heat capacity through inquiry

Ontario Grade 11 Physics

Energy and Society

Planning measurements, using energy relationships, and judging evidence

When two materials receive the same amount of energy, their temperatures may change by different amounts. Specific heat capacity describes how much energy is needed to change the temperature of a given mass of a material. In this lesson, the system is the sample being heated. We choose energy transferred into the sample as positive; energy leaving it is negative. Heat and temperature change are scalar quantities, so they have magnitude but no direction. The sign convention still helps us describe whether energy enters or leaves the sample.

What you will learn

Describe specific heat capacity and state its SI unit.
Plan an investigation that uses measured mass and temperature change to determine specific heat capacity.
Use an energy relationship to calculate specific heat capacity and check whether a result is reasonable.
Identify measurement limits and ways to improve an investigation.

1. Bridge from mass, temperature, and energy

Mass is the amount of matter in a sample. In this investigation, measure mass in kilograms. Temperature tells how hot or cold a sample is. Record temperature in degrees Celsius or kelvins, but use a temperature change in degrees Celsius or kelvins. A change of one degree Celsius has the same size as a change of one kelvin.

Energy is measured in joules. Specific heat capacity, represented by

c

, is the energy needed to raise the temperature of one kilogram of a material by one degree Celsius. Its SI unit is joules per kilogram per degree Celsius. A high specific heat capacity means that more energy is needed for the same temperature rise in the same mass.

The temperature change is final temperature minus initial temperature. If the sample warms, this change is positive. If it cools, it is negative. In the heating relationship used here,

Q

is energy transferred into the sample,

m

is its mass, and

\Delta T

is its temperature change. The relationship assumes that the measured energy transfer goes into changing the sample's temperature.

Q=mc\Delta T

Specific heat capacity is a property of a material.
Use kilograms, joules, and degrees Celsius or kelvins for temperature change.
A warmer sample has a positive temperature change.

2. Design an inquiry that can determine c

An inquiry is a planned investigation that uses evidence to answer a question. Here, the question could be: What is the specific heat capacity of this sample? To answer it, measure the sample's mass, its initial temperature, its final temperature, and the energy transferred to it. Rearrange the relationship to calculate

c

.

One possible procedure uses an electric heater in contact with a sample. Measure the sample's mass with a balance and its initial temperature with a thermometer. Supply energy for a measured time. If the heater's power is known, calculate supplied energy from power multiplied by time. Power is measured in watts, and one watt is one joule per second. Record the final temperature, then find the temperature change.

This is a proposed procedure, not a report of completed measurements. In a real investigation, the readings from the balance, thermometer, timer, and heater are measured evidence. A calculated result is based on that evidence; it is not itself a direct measurement of specific heat capacity.

Not all supplied energy necessarily warms the sample. Some can warm the container or escape to the surroundings. A lid and insulation can reduce energy loss. The sample should be stirred gently, if appropriate, so its temperature is more even. Repeat trials and compare the results. Record instrument precision and any uncertainty, meaning the range within which a measurement is likely to lie.

c=\frac{Q}{m\Delta T}

Measure mass, starting and ending temperatures, and energy supplied.
Use repeated trials to check consistency.
Energy loss and heating the container can affect the result.

3. Read the result as evidence

A value calculated from one set of readings is an estimate. Compare repeated values and check whether differences could come from reading precision or energy loss. If the sample's temperature rise is very small, thermometer resolution can make the calculated value less reliable. If the sample loses energy to the room, using all supplied energy as though it entered the sample tends to make the calculated

c

too large.

Keep units in the calculation. In the expression for

c

, joules are divided by kilograms and degrees Celsius. The result therefore has units of joules per kilogram per degree Celsius. Report a sensible number of significant figures, which are the meaningful digits supported by the measurements.

The simple model is most useful when the sample's temperature is fairly uniform and the measured energy transfer is close to the energy absorbed by the sample. Explain limitations rather than pretending the measurement is exact. A simulation can help practise calculations, but simulated values are not experimental evidence.

1\ \mathrm{W}=1\ \mathrm{J/s}

A calculated value depends on the measurements and the model's assumptions.
Check units and whether the value is physically plausible for the material.
State limitations such as heat loss and instrument resolution.

Worked example

Calculate c from a measured energy transfer

A proposed investigation would transfer

8400\ \mathrm{J}

into a

0.500\ \mathrm{kg}

sample. Its temperature would rise from

20.0\ ^\circ\mathrm{C}

to

28.0\ ^\circ\mathrm{C}

. Determine the sample's estimated specific heat capacity.

Set the system and direction
The system is the sample. Energy enters it, so $Q$ is positive. The sample warms, so its temperature change is positive. The unknown is $c$ .
$Q=+8400\ \mathrm{J},\quad \Delta T=28.0-20.0=+8.0\ ^\circ\mathrm{C}$
Choose the relationship
Use the energy relationship and solve for specific heat capacity. The temperature difference is expressed in degrees Celsius.
$c=\frac{Q}{m\Delta T}$
Substitute and calculate
Substitute the measured values with their units. Keep the positive sign because both energy transfer into the sample and its warming are positive.
$c=\frac{8400\ \mathrm{J}}{(0.500\ \mathrm{kg})(8.0\ ^\circ\mathrm{C})}=2.1\times10^3\ \mathrm{J/(kg\cdot{}^\circ C)}$

Answer: The estimated specific heat capacity is

2.1\times10^3\ \mathrm{J/(kg\cdot{}^\circ C)}

, to two significant figures.

Check: The units reduce to joules per kilogram per degree Celsius. A positive value is reasonable because energy warmed the sample. The value is an estimate; energy lost to the surroundings would affect it.

Worked example

Find specific heat capacity from heater power and time

In a proposed setup, a heater rated at

30.0\ \mathrm{W}

runs for

120\ \mathrm{s}

while heating a

0.200\ \mathrm{kg}

sample. The sample warms by

18.0\ ^\circ\mathrm{C}

. Assuming all heater energy enters the sample, determine

c

.

Define the system and direction
The system is the sample, and the positive energy direction is into it. Heater power is energy transferred each second. The unknown is the sample's specific heat capacity.
$Q=Pt$
Calculate the supplied energy
Multiply power by time. Watts are joules per second, so multiplying by seconds gives joules.
$Q=(30.0\ \mathrm{J/s})(120\ \mathrm{s})=3600\ \mathrm{J}$
Calculate c
Use the measured mass and temperature rise. The energy and temperature change are both positive, so the calculated value is positive.
$c=\frac{3600\ \mathrm{J}}{(0.200\ \mathrm{kg})(18.0\ ^\circ\mathrm{C})}=1.00\times10^3\ \mathrm{J/(kg\cdot{}^\circ C)}$

Answer: Under the stated assumption, the estimated specific heat capacity is

1.00\times10^3\ \mathrm{J/(kg\cdot{}^\circ C)}

.

Check: The units are correct because joules are divided by kilograms and degrees Celsius. The positive result matches the warming. In a real setup, some heater energy may warm the container or escape, so the assumption should be checked.

Worked example

Account for cooling and the sign convention

A

0.250\ \mathrm{kg}

sample cools from

65.0\ ^\circ\mathrm{C}

to

53.0\ ^\circ\mathrm{C}

. During this change, it transfers

3000\ \mathrm{J}

of energy out to its surroundings. Determine its estimated specific heat capacity.

Define the system and signs
The system is the sample. Energy leaving the sample is negative, and cooling gives a negative temperature change. The two negative signs produce a positive specific heat capacity.
$Q=-3000\ \mathrm{J},\quad \Delta T=53.0-65.0=-12.0\ ^\circ\mathrm{C}$
Apply the relationship
Use the same energy relationship for cooling as for warming. The signs show that the sample loses energy as its temperature falls.
$c=\frac{Q}{m\Delta T}$
Substitute and check the value
Substitute both negative quantities. Keep sufficient digits during calculation, then report the result to three significant figures.
$c=\frac{-3000\ \mathrm{J}}{(0.250\ \mathrm{kg})(-12.0\ ^\circ\mathrm{C})}=1.00\times10^3\ \mathrm{J/(kg\cdot{}^\circ C)}$

Answer: The estimated specific heat capacity is

1.00\times10^3\ \mathrm{J/(kg\cdot{}^\circ C)}

.

Check: The signs cancel, giving a positive material property. The units are joules per kilogram per degree Celsius. The energy loss and temperature fall agree in direction.

Common mistakes and how to avoid them

Using final temperature as the temperature change.

Correction: Subtract initial temperature from final temperature. Use only the difference in the relationship.

Reporting specific heat capacity in joules alone.

Correction: Divide energy by both mass and temperature change. The unit is

\mathrm{J/(kg\cdot{}^\circ C)}

.

Treating all heater energy as energy absorbed by the sample without stating the assumption.

Correction: State that assumption and explain that energy loss or heating the container can affect the estimate.

Making the energy and temperature-change signs disagree during cooling.

Correction: With the sample as the system, energy leaving and cooling are both negative.

Lesson summary

Specific heat capacity describes the energy required per kilogram per degree of temperature change.
Use $Q=mc\Delta T$ and rearrange it to determine $c$ .
A suitable inquiry measures mass, temperature change, and energy transferred.
Report units and significant figures, then discuss signs, reasonableness, and sources of uncertainty.

Check your understanding

Question 1

A sample of mass

0.400\ \mathrm{kg}

absorbs

4800\ \mathrm{J}

and warms by

10.0\ ^\circ\mathrm{C}

. What is its estimated specific heat capacity?

$1200\ \mathrm{J/(kg\cdot{}^\circ C)}$
$120\ \mathrm{J/(kg\cdot{}^\circ C)}$
$12\ \mathrm{J/(kg\cdot{}^\circ C)}$
$1200\ \mathrm{J}$

Show answer and explanation

1200\ \mathrm{J/(kg\cdot{}^\circ C)}

Using

c=Q/(m\Delta T)

gives

4800/(0.400\times10.0)=1200

. The units must include kilograms and degrees Celsius.

Question 2

For the sample as the system, which signs describe energy leaving while the sample cools?

$Q<0$ and $\Delta T<0$
$Q>0$ and $\Delta T<0$
$Q<0$ and $\Delta T>0$
$Q>0$ and $\Delta T>0$

Show answer and explanation

Q<0

and

\Delta T<0

Energy leaving the chosen system is negative. Cooling means final temperature is below initial temperature, so the temperature change is negative.

Question 3

Why should an investigation use insulation around the sample?

To reduce energy transfer between the sample and its surroundings
To increase the sample's mass without measuring it
To make the temperature change negative
To remove the need to measure the initial temperature

Show answer and explanation

To reduce energy transfer between the sample and its surroundings

Insulation reduces energy loss to or gain from the surroundings. It helps the measured energy transfer better represent the energy entering the sample.

Key terms

Specific heat capacity: The energy needed to change the temperature of one kilogram of a material by one degree Celsius or one kelvin.
Temperature change: Final temperature minus initial temperature.
Inquiry: A planned investigation that gathers evidence to answer a question.
Uncertainty: The range around a measured value that reflects the limits of the measuring process.
Significant figures: Digits that show the meaningful precision of a measured or calculated value.

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About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Ontario Grade 11 Physics (SPH3U), expectation D2.9. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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