Calorimetry | ICSE Class 10 Physics Notes

This note covers heat and temperature, thermal capacity, specific heat capacity, calorimeters, the principle of mixtures, water’s high specific heat capacity, changes of state, heating curves, latent heat of fusion, and calculations involving warming, cooling, melting and freezing.

What are heat, temperature and calorimetry?

Heat is energy transferred between bodies, or between a body and its surroundings, because their temperatures differ. Temperature measures relative hotness or coldness. A temperature difference determines the direction of heat transfer: energy passes from the hotter part to the colder part.

A glass of ice-cold water on a table gains heat from warmer surroundings. Hot tea on the same table loses heat to those surroundings. The transfer continues until the body and its surroundings reach the same temperature.

Definition: Calorimetry means measurement of heat. A calorimeter is a device in which heat measurements can be made.

Which units describe these quantities?

The International System of Units, abbreviated SI, provides standard units. The SI unit of heat is the joule, symbol J. The SI unit of temperature is the kelvin, symbol K. Degree Celsius, written °C, is another commonly used temperature unit.

A calorie, symbol cal, is the heat required to raise the temperature of one gram of water from 14.5 °C to 15.5 °C. The gram, symbol g, is a mass unit; the SI unit of mass is the kilogram, symbol kg.

1 cal = 4.186 J

Let T denote temperature in kelvin and t temperature in degrees Celsius. Their relationship is:

T = t + 273.15

The two scales have equal-sized temperature intervals. Consequently, a temperature rise expressed in kelvin has the same numerical value as the rise expressed in degrees Celsius. The offset is used when converting a temperature, not when converting a temperature difference.

Note: Heating need not produce a temperature rise. Energy supplied during melting can change the state while the temperature remains constant. Temperature change alone therefore does not describe every effect of supplying heat.

How do heat capacity and specific heat capacity differ?

Heat capacity, also called thermal capacity, describes a whole body. It is the heat required per unit rise in that body’s temperature, provided no change of state occurs. A change of state means a transition between solid, liquid and gas.

Let Q be the heat supplied, ΔT the resulting temperature rise, and C′ the body’s heat capacity. The symbol Δ means “change in”. For a temperature rise, subtract the initial temperature from the final temperature.

$C' = Q/\Delta T$

The SI unit of heat capacity is J K⁻¹, read as joules per kelvin. The superscript ⁻¹ means “per”. A larger heat capacity means that more heat is needed for the same temperature rise of the whole body.

Specific heat capacity is the heat required per unit mass per unit temperature rise, without a change of state. Let m be the mass and C the specific heat capacity of its material.

C = Q/(mΔT)

The SI unit of specific heat capacity is J kg⁻¹ K⁻¹, meaning joules per kilogram per kelvin. Specific heat capacity depends on the substance and its temperature; it is not simply a measure of the quantity of material present.

Derivation: How are the two capacities related?

1. For the whole body, heat capacity is the supplied heat divided by its temperature rise: C′ = Q/ΔT.
2. For unit mass of the same material, specific heat capacity is C = Q/(mΔT).
3. Multiplying the second expression by mass gives mC = Q/ΔT, which equals the body’s heat capacity.

$C' = mC$

What determines the heat needed to change temperature?

For a substance undergoing no change of state, the heat required depends on its mass, its temperature change and its specific heat capacity. Rearranging the definition of specific heat capacity gives the main relationship used for warming and cooling.

Q = mCΔT

Use a specific heat capacity appropriate to the substance and the stated conditions. In simple calculations it is treated as constant over the temperature interval. The formula must not be used as the energy required for melting at constant temperature.

How can the three factors be compared?

Heating a given amount of water through 40 °C takes about twice the time needed for a 20 °C rise with the same heat source. The associated heat requirement is doubled. Increasing the mass also increases the heat needed for a specified rise.

Replacing water with the same mass of mustard oil changes the result. For the same 20 °C rise, the oil needs less heat. Equal masses of different substances do not necessarily undergo equal temperature rises when they receive equal quantities of heat.

For an unknown temperature rise, rearrange the relationship:

ΔT = Q/(mC)

For equal masses receiving equal heat, a higher specific heat capacity gives a smaller temperature rise. This conclusion compares the same energy input and assumes that neither substance changes state. Without those conditions, the comparison is incomplete.

Note: In a heat-loss calculation, use the initial temperature minus the final temperature to obtain the positive temperature fall. In a heat-gain calculation, use the final temperature minus the initial temperature to obtain the positive rise.

Why is water’s high specific heat capacity important?

Water has a high specific heat capacity compared with many familiar materials. A given mass can absorb considerable heat with a comparatively small temperature rise. It can also release considerable heat as it cools through a given temperature interval.

Which values should be compared?

These are reference values; specific heat capacity depends on temperature. Ice and liquid water have different values because their physical states differ. In calculations, use the value supplied for that particular problem, including its stated rounding.

Water serves as a coolant, a substance used to absorb and carry away heat, in automobile radiators. Its high specific heat capacity allows substantial heat absorption. Water in a hot-water bag can give out substantial heat while its temperature falls.

How does this affect land and sea?

During summer, water warms more slowly than land, and wind from the sea has a cooling effect. During the day, the ground heats more quickly than large bodies of water. Mixing currents also distribute absorbed heat through a large volume of water.

Air warmed near the land expands, becomes less dense and rises. Density is mass per unit volume. Cooler air moves in from over the water, producing a sea breeze. At night, land loses heat more quickly and the water surface is warmer, so the circulation reverses.

Convection means heat transfer by the actual movement of matter in a liquid or gas. The breeze involves convection, while water’s high specific heat capacity helps explain the different temperature responses of land and water. Both ideas contribute to the explanation.

How does a calorimeter help measure heat?

A calorimeter allows heat exchanged between its contents to be related to measured temperature changes. A simple form contains a metallic vessel and a stirrer made of the same material, such as copper or aluminium. The stirrer is used to mix the contents.

The vessel sits inside a wooden jacket containing insulating material, such as glass wool. Insulation reduces heat transfer between the contents and the surroundings. An opening allows a thermometer, an instrument for measuring temperature, to enter the calorimeter.

Draw an inner metallic vessel with a stirrer and thermometer entering its contents. Surround it with an outer wooden jacket and label the insulating glass wool between the inner vessel and the jacket.

What is the condition for a reliable heat balance?

For the heat balance used here, a thermally isolated system exchanges no heat with its surroundings. Insulation reduces unwanted transfer, while the ideal calculation assumes that the heat exchange with the surroundings is zero.

If a hot solid is placed in cooler water, both the water and the calorimeter can gain heat. The vessel must therefore be included when its mass and specific heat capacity are supplied. It is part of the thermal system being considered.

1. Identify the hot body, the cooler contents and the calorimeter, together with their masses and initial temperatures.
2. Bring the bodies into thermal contact, meaning contact that allows heat transfer, and mix the contents using the stirrer.
3. Record the common final temperature reached by the contents and the calorimeter.
4. Calculate the heat lost and gained, including the vessel when its heat capacity is relevant.

The common-temperature condition is called thermal equilibrium. A measured final temperature supplies the temperature changes used in the energy balance. Insulation does not justify silently omitting the calorimeter’s own heat gain.

How is the principle of mixtures applied?

Definition: The principle of mixtures states that heat lost by the hotter parts equals heat gained by the colder parts, provided no heat is exchanged with the surroundings.

This is an application of the law of conservation of energy: energy is transferred between parts of the system, rather than created or destroyed. A calculation must include every body that loses or gains an appreciable quantity of heat.

Derivation: What is the mathematical statement for two bodies?

Let m₁ and C₁ be the hot body’s mass and specific heat capacity, and m₂ and C₂ the corresponding quantities for the cold body. Let T₁ and T₂ be their initial temperatures, and T be their common final temperature.

1. The hot body cools through T₁ − T, so its heat loss is m₁C₁(T₁ − T).
2. The cold body warms through T − T₂, so its heat gain is m₂C₂(T − T₂).
3. Equate these quantities when the two bodies exchange no heat with the surroundings, no state changes occur, and any vessel’s heat exchange is negligible.

m₁C₁(T₁ − T) = m₂C₂(T − T₂)

When the calorimeter also warms, add its heat gain to the colder side. Do not average the initial temperatures without accounting for the masses and specific heat capacities that determine each body’s heat exchange.

Worked example 1. A 0.047 kg aluminium sphere at 100 °C enters a 0.14 kg copper calorimeter containing 0.25 kg water at 20 °C. Their final temperature is 23 °C. Find the sphere’s specific heat capacity. Use water: 4180 J kg⁻¹ K⁻¹; copper: 386 J kg⁻¹ K⁻¹. Neglect external heat exchange.

Formula: Q₁ = m₁C₁ΔT₁; Q₂ = m₂C₂ΔT₂. Here Q₁ is the sphere’s heat loss; Q₂ is the water’s heat gain; ΔT₁ and ΔT₂ are their positive temperature changes. Add the calorimeter’s heat gain separately.

Substitute: sphere’s fall = 100 − 23 = 77 K; water and vessel’s rise = 23 − 20 = 3 K. Thus 0.047 × C₁ × 77 = (0.25 × 4180 + 0.14 × 386) × 3.

Answer: C₁ = 3297.12/3.619 ≈ 911 J kg⁻¹ K⁻¹. Including the copper vessel accounts for the heat it absorbs while warming with the water.

What happens during melting and freezing?

Matter normally exists as solid, liquid or gas. A transition between these states is a change of state. Melting, or fusion, changes a solid into a liquid; freezing changes a liquid into a solid.

Pressure is force per unit area. During melting at fixed pressure, solid and liquid coexist at the same temperature. Heat continues to enter, but it changes the state instead of increasing the temperature. The temperature remains constant until the entire solid has melted.

The melting point is the temperature at which solid and liquid coexist in thermal equilibrium. It depends on the substance and on pressure. The melting point at standard atmospheric pressure, the reference pressure of one atmosphere, is called the normal melting point.

How can pressure affect melting ice?

Increased pressure can lower the melting temperature of ice. A weighted metallic wire placed over an ice slab can pass through it because ice melts under the increased pressure below the wire. Above the wire, the water freezes again.

This refreezing is called regelation. The wire passes through without splitting the slab into separate pieces. The process connects a pressure-dependent melting point with successive melting and freezing, rather than a continuous rise in the slab’s temperature.

The drawing shows an ice slab resting across two supports. A wire runs over the slab, with weights hanging from its ends. A dotted part indicates the wire’s path inside the slab.

See Fig. 10.10 in your NCERT textbook

Distinguish the state change from ordinary cooling. Water can lose heat while freezing without falling below its freezing temperature during that change. Likewise, ice at its melting point can absorb heat without warming while some ice remains.

How should the heating curve for water be read?

A heating curve shows how temperature changes as a substance receives heat. A phase means a physical state here. A graph can use elapsed time on the horizontal axis when heating proceeds with a constant source, or use heat supplied directly on that axis.

For an ice-and-water mixture, the temperature stays steady while ice melts. After all the ice has become water, further heating raises the temperature. The water reaches nearly 100 °C before its temperature becomes steady again as it changes into vapour, the gaseous state, under ordinary atmospheric conditions. Steam is water vapour.

The vertical axis is temperature in °C and the horizontal axis is time in minutes. A horizontal segment at 0 °C is followed by a rising segment and a horizontal segment at 100 °C. A dotted rising extension follows. The graph is not to scale.

See Fig. 10.9 in your NCERT textbook

What do rising and horizontal parts mean?

Boiling is the conversion of liquid into vapour at its boiling point, the temperature where liquid and vapour coexist. The boiling point depends on pressure. The conventional 100 °C label describes the familiar atmospheric-pressure heating curve, rather than every possible boiling condition.

At a pressure of one atmosphere, the graph labels solid ice, liquid water and steam. Rising portions are separated by shaded phase-change regions at the marked melting and boiling points. The horizontal axis is heat and the vertical axis is temperature. It is not to scale.

See Fig. 10.12 in your NCERT textbook

The different slopes of the rising portions show that the specific heat capacities of the different states are not equal. A horizontal part shows continuing energy transfer without a temperature change.

What is specific latent heat of fusion?

Latent heat refers to energy absorbed or released during a change of state without a temperature change. To distinguish a total energy from a material property, use specific latent heat for the energy transferred per unit mass.

The specific latent heat of fusion is the heat required per unit mass to change a solid into a liquid at the same temperature and pressure. Let L denote this quantity for fusion. Then:

L = Q/m

Q = mL

The SI unit of specific latent heat of fusion is J kg⁻¹, or joules per kilogram. Total latent heat is an energy and is expressed in joules. A kelvin does not appear in the specific latent heat unit because there is no temperature rise in this process.

What does the value for ice mean?

A reference value for ice is 3.33 × 10⁵ J kg⁻¹. It means that melting one kilogram of ice at its melting point requires 3.33 × 10⁵ J without a temperature increase. Latent heat depends on pressure and is usually quoted at standard atmospheric pressure.

The reverse change releases heat. When water freezes, it transfers the corresponding fusion energy to its surroundings. Keep the direction clear: melting absorbs energy; freezing releases energy. At the transition temperature, both changes can occur without a change in temperature.

Some problems use rounded fusion values such as 3.35 × 10⁵ J kg⁻¹. Use the stated value throughout that problem. Changing to another rounded reference value halfway through a calculation can change the numerical result.

How are warming and melting calculations separated into stages?

If ice starts below its melting point, supplying heat first raises the ice temperature. Further heat melts it, and after melting is complete, additional heat raises the water temperature. Each stage requires its own expression because the physical process changes.

How does a stage-by-stage calculation work?

1. Write the starting temperature and state, then the required final temperature and state.
2. Use the specific heat capacity of ice for warming solid ice to its melting point.
3. Use the specific latent heat of fusion for the melting stage at constant temperature.
4. Use the specific heat capacity of water for warming the liquid, then add the required stage energies.

Worked example 2. Find the heat needed to warm 3 kg of ice from −12 °C to 0 °C without melting. Use the ice’s specific heat capacity of 2100 J kg⁻¹ K⁻¹.

Formula: Q = mCΔT. Substitute: the rise is 0 − (−12) = 12 K, so Q = 3 × 2100 × 12.

Answer: 75,600 J. The ice has reached its melting point; this calculation has not yet supplied the energy needed to turn it into water.

Worked example 3. Melt 3 kg of ice at 0 °C into water at 0 °C. Take the specific latent heat of fusion as 3.35 × 10⁵ J kg⁻¹.

Formula: Q = mL. Substitute: Q = 3 × 3.35 × 10⁵.

Answer: 1,005,000 J. The temperature does not rise, although a substantial quantity of heat is supplied. The solid becomes liquid during this stage.

Worked example 4. Warm 3 kg of liquid water from 0 °C to 100 °C without converting it into steam. Use a specific heat capacity of 4186 J kg⁻¹ K⁻¹.

Formula: Q = mCΔT. Substitute: Q = 3 × 4186 × (100 − 0).

Answer: 1,255,800 J. This energy warms liquid water; it does not include energy for converting the water at its final temperature into vapour.

Combining these three stages takes 3 kg of ice at −12 °C to water at 100 °C. The required heat is 75,600 + 1,005,000 + 1,255,800 = 2,336,400 J. This sum assumes that the supplied energy goes into the ice and water.

Notice why a single temperature-change calculation fails: it would overlook the energy absorbed during melting. It would also apply one specific heat capacity across both solid and liquid states, even though the two given values are different.

How do mixtures involving ice combine heat loss and fusion?

When a hot body meets ice, its heat loss can supply energy for melting. If the resulting water ends above the melting point, that water must also gain heat after melting. List these separate requirements before writing the heat balance.

How much ice can a hot copper block melt?

Worked example 5. A 2.5 kg copper block at 500 °C is placed on a large ice block at 0 °C. Find the maximum mass melted. Use copper’s specific heat capacity of 390 J kg⁻¹ K⁻¹ and fusion energy of 335,000 J kg⁻¹. Assume all the released heat melts ice and the copper cools to 0 °C.

Formula: Q = mCΔT; M = Q/L. Here M denotes the mass of ice melted, while m is the copper mass.

Substitute: Q = 2.5 × 390 × 500 = 487,500 J; M = 487,500/335,000.

Answer: approximately 1.46 kg of ice. This is a maximum because the stated assumption allocates all the copper’s released heat to fusion.

How is fusion energy found from a mixture?

Worked example 6. Mix 0.15 kg of ice at 0 °C with 0.30 kg of water at 50 °C. All the ice melts and the final temperature is 6.7 °C. Find the specific latent heat of fusion. Use water’s specific heat capacity of 4186 J kg⁻¹ K⁻¹; neglect the container’s heat capacity and external heat exchange.

Formula: Q₁ = m₁CΔT₁; Q₂ = m₂CΔT₂. Here Q₁ is the original warm water’s heat loss and Q₂ is the melted ice water’s heat gain; m₁ and m₂ are their respective masses.

Substitute: Q₁ = 0.30 × 4186 × (50 − 6.7) = 54,376.14 J. Q₂ = 0.15 × 4186 × 6.7 = 4206.93 J.

Answer: 54,376.14 = 0.15L + 4206.93, so L = (54,376.14 − 4206.93)/0.15 ≈ 3.34 × 10⁵ J kg⁻¹.

The second example includes two heat gains: melting the ice and warming the melted water. Assigning all the warm water’s heat loss to fusion would give too large a value for the specific latent heat.

Check the final state before accepting any answer. If the energy available is insufficient to melt all the ice, a calculation that assumes all the resulting water warms above the melting point is inconsistent with its own energy balance.

Glossary

• Heat — Energy transferred between bodies or systems because their temperatures differ.
• Temperature — A measure of relative hotness or coldness that determines the direction of heat transfer.
• Calorimetry — Measurement of heat exchanged during processes such as warming, cooling and melting.
• Calorimeter — A device used for heat measurements through controlled heat exchange and temperature observations.
• Heat capacity — Heat required for a unit temperature rise of an entire body without a state change.
• Specific heat capacity — Heat required per unit mass for a unit temperature rise without a state change.
• Thermal equilibrium — The condition in which bodies in thermal contact have reached the same temperature.
• Thermal insulation — Material or an arrangement that reduces heat transfer between a system and its surroundings.
• Fusion — The change of a substance from its solid state into its liquid state.
• Melting point — Temperature at which a substance’s solid and liquid states coexist in thermal equilibrium at a specified pressure.
• Specific latent heat of fusion — Heat required per unit mass to change a solid into liquid at constant temperature and pressure.
• Freezing — The change from liquid to solid, accompanied by release of heat during the transition.
• Regelation — Refreezing of water after ice melts under increased pressure and that pressure is removed.
• Heating curve — A graph showing temperature during heating against time or the quantity of heat supplied.

Common errors and misconceptions

• Misconception: Heat and temperature are interchangeable quantities. Correct: Heat is transferred energy, measured in joules; temperature measures hotness or coldness and has kelvin as its SI unit.
• Misconception: Heat capacity and specific heat capacity have the same unit. Correct: Their SI units are J K⁻¹ and J kg⁻¹ K⁻¹ respectively, because the second quantity is defined per unit mass.
• Misconception: Every temperature difference in Celsius needs 273.15 added. Correct: Celsius and kelvin intervals have equal size. The offset converts a temperature, not a temperature rise or fall.
• Misconception: The calorimeter absorbs no heat because it is insulated. Correct: Insulation reduces external exchange. The calorimeter can still warm with its contents and must be included when relevant.
• Misconception: Melting needs no heat because the temperature stays constant. Correct: Fusion requires energy even without a temperature rise; calculate it using mass multiplied by specific latent heat.
• Misconception: All heat lost by warm water goes into melting ice. Correct: If the final temperature exceeds the melting point, some energy also warms the water produced from the ice.
• Misconception: Water has the highest specific heat capacity of every substance. Correct: Its specific heat capacity is high compared with the familiar substances compared here; that comparison does not establish a universal maximum.
• Misconception: Every substance melts or boils at a pressure-independent temperature. Correct: Transition temperatures depend on pressure, so the pressure condition matters when interpreting a heating curve.

Exam-style questions with model answers

Q1. Define calorimetry and state the SI unit of heat. [2 marks]

1. Calorimetry is the measurement of heat exchanged during a physical process.
2. The SI unit of heat is the joule, represented by the symbol J.

Q2. Distinguish heat capacity from specific heat capacity by definition, SI unit and their relationship. [3 marks]

1. Heat capacity describes the heat needed for a unit temperature rise of the whole body. Specific heat capacity describes the corresponding requirement per unit mass, without a change of state.
2. The SI unit of heat capacity is J K⁻¹, whereas that of specific heat capacity is J kg⁻¹ K⁻¹.
3. The relationship is C′ = mC, where C′ is heat capacity, m is mass and C is specific heat capacity.

Q3. Why is water useful in an automobile radiator and a hot-water bag? Relate both uses to its specific heat capacity. [3 marks]

1. Water has a high specific heat capacity, so a given mass exchanges a comparatively large quantity of heat for a given temperature change.
2. In an automobile radiator, this enables water to absorb substantial heat while its own temperature rise remains comparatively small.
3. In a hot-water bag, water gives out substantial heat as it cools, providing heat to its surroundings during its temperature fall.

Q4. Calculate the heat needed to warm 3 kg of ice from −12 °C to 0 °C without melting. Its specific heat capacity is 2100 J kg⁻¹ K⁻¹. Explain why latent heat is not included. [4 marks]

1. The temperature rise is 0 − (−12) = 12 °C, equal in size to 12 K.
2. Use Q = mCΔT, where Q is heat, m is mass, C is specific heat capacity and ΔT is temperature rise.
3. Substitution gives Q = 3 × 2100 × 12 = 75,600 J of heat supplied to the ice.
4. No latent heat term is included because the stated process ends with solid ice at its melting point, without melting it.

Q5. A 0.047 kg aluminium sphere at 100 °C enters a 0.14 kg copper calorimeter holding 0.25 kg water, both initially at 20 °C. The common final temperature is 23 °C. Calculate the sphere’s specific heat capacity. Use water: 4180 J kg⁻¹ K⁻¹ and copper: 386 J kg⁻¹ K⁻¹. Assume no heat exchange with the surroundings and no state change. [5 marks]

1. Apply the principle of mixtures: the aluminium sphere’s heat loss equals the combined heat gain of the water and copper calorimeter, since the system exchanges no heat with its surroundings.
2. The sphere’s temperature fall is 100 − 23 = 77 K. The water and copper vessel each rise through 23 − 20 = 3 K.
3. The water gains 0.25 × 4180 × 3 = 3135 J, while the copper vessel gains 0.14 × 386 × 3 = 162.12 J.
4. Let C be the sphere’s specific heat capacity. The heat balance becomes 0.047 × C × 77 = 3135 + 162.12 = 3297.12 J.
5. Therefore C = 3297.12/(0.047 × 77) ≈ 911 J kg⁻¹ K⁻¹. The vessel’s heat gain is included because it warms with the water.

Q6. Describe and explain the melting and liquid-warming portions of the heating curve when ice at 0 °C is heated at constant atmospheric pressure. [3 marks]

1. During melting, the curve has a horizontal portion: solid ice and liquid water coexist at the same temperature while heat continues to enter.
2. The supplied energy changes the state from solid to liquid, rather than increasing the temperature. This continues until all the ice melts.
3. Once melting is complete, further heating raises the temperature of the liquid water, producing a rising portion of the curve before boiling begins.

Q7. Find the heat needed to convert 3 kg of ice at −12 °C into water at 100 °C at atmospheric pressure, without forming steam. Use ice’s specific heat capacity 2100 J kg⁻¹ K⁻¹, water’s specific heat capacity 4186 J kg⁻¹ K⁻¹, melting point 0 °C and specific latent heat of fusion 335,000 J kg⁻¹. Neglect the vessel’s heat capacity and external losses. [5 marks]

1. Separate the process into warming solid ice, melting it at its melting point and warming the resulting liquid. No energy for producing steam is required by the stated final state.
2. For warming the ice, use heat = mass × specific heat capacity × temperature rise. The heat required is 3 × 2100 × [0 − (−12)] = 75,600 J.
3. For melting, use heat = mass × specific latent heat of fusion. This stage requires 3 × 335,000 = 1,005,000 J at constant temperature.
4. For warming the liquid water, use its own specific heat capacity. The heat required is 3 × 4186 × (100 − 0) = 1,255,800 J.
5. Add the three energy requirements: total heat = 75,600 + 1,005,000 + 1,255,800 = 2,336,400 J, under the stated assumptions about heat losses and the vessel.

Q8. Mix 0.15 kg of ice at its melting point of 0 °C with 0.30 kg of water at 50 °C. All the ice melts; the final temperature is 6.7 °C. Calculate the specific latent heat of fusion, using water’s specific heat capacity of 4186 J kg⁻¹ K⁻¹. Neglect the container’s heat capacity and heat exchange with the surroundings. [5 marks]

1. The original warm water supplies the heat needed both to melt the ice and to warm the resulting water. These are separate heat gains in the energy balance.
2. The original water’s temperature fall is 50 − 6.7 = 43.3 K. Its heat loss is 0.30 × 4186 × 43.3 = 54,376.14 J.
3. The water produced from the ice warms through 6.7 K after melting. It gains 0.15 × 4186 × 6.7 = 4206.93 J during that warming stage.
4. The energy available specifically for melting is therefore 54,376.14 − 4206.93 = 50,169.21 J. Divide this energy by the mass of ice that melts.
5. The specific latent heat of fusion is 50,169.21/0.15 ≈ 3.34 × 10⁵ J kg⁻¹. The result includes the liquid-water warming requirement rather than assigning all heat loss to fusion.

Key takeaways

• Heat is energy transferred because of a temperature difference; temperature measures hotness or coldness.
• Heat capacity describes a whole body, while specific heat capacity gives the heat requirement per unit mass and temperature rise.
• Use Q = mCΔT for warming or cooling without a state change, choosing the correct material’s specific heat capacity.
• The principle of mixtures equates heat loss and heat gain when the system exchanges no heat with its surroundings.
• Water’s high specific heat capacity helps explain its use in radiators and hot-water bags, and the differing temperature responses of land and sea.
• Melting absorbs heat and freezing releases heat while temperature remains constant during the state change at fixed pressure.
• Specific latent heat of fusion is energy per unit mass, measured in J kg⁻¹; total fusion energy equals mass multiplied by this value.
• Split ice problems into the stages actually required: warming ice, melting ice and, when necessary, warming the resulting water.

Test yourself

What makes heat transfer occur between two bodies?

A temperature difference causes heat transfer from the hotter body to the colder body.

Why does a Celsius temperature rise have the same numerical value in kelvin?

The Celsius and kelvin scales have equal-sized intervals, although their zero points differ.

How is a body’s heat capacity related to its mass and specific heat capacity?

Heat capacity equals mass multiplied by specific heat capacity for the material under the stated conditions.

Why must a copper calorimeter sometimes appear in the heat balance?

It can warm with its contents, so it absorbs some of the heat released by the hotter body.

What does the horizontal melting portion of a heating curve show?

Heat changes solid into liquid while the temperature stays constant and both states coexist.

Which quantity is measured in J kg⁻¹: specific heat capacity or specific latent heat?

Specific latent heat is measured in J kg⁻¹; specific heat capacity also includes a per-kelvin factor.

What extra energy term is needed if melted ice finishes above its melting point?

Include the heat required to warm the resulting water from the melting point to the final temperature.

Does freezing absorb or release the fusion energy?

Freezing releases heat to the surroundings as liquid changes into solid at the transition temperature.