Chapter 9 Heat
Class 10 Science Notes
Table of Contents (Chapter 9 Heat)
1. Summary: Heat
1.1 Important Terms & Definitions
Heat
Definition: A form of energy that gives the sensation of warmth. It is the total kinetic energy of the molecules of a substance. It flows from a body at a higher temperature to a body at a lower temperature.
Temperature
Definition: The degree of hotness or coldness of a body. It is the measure of the average kinetic energy of the molecules.
Specific Heat Capacity ($s$)
Definition: The amount of heat energy required to change the temperature of $1 \text{kg}$ of a substance by $1^\circ\text{C}$ (or $1 \text{K}$).
Unit: Joule per kilogram degree Celsius ($J/kg^\circ C$).
Thermal Equilibrium
Definition: A state reached when two bodies in contact are at the same temperature and there is no net flow of heat between them.
Calorimetry
Definition: The branch of physics that deals with the measurement of heat.
Anomalous Expansion of Water
Definition: The unusual behavior of water where it expands instead of contracting when cooled from $4^\circ\text{C}$ to $0^\circ\text{C}$.
1.2 Fundamental Principle
The Principle of Calorimetry (Principle of Heat Exchange)
When two bodies of different temperatures are kept in thermal contact, heat flows from the hot body to the cold body until they reach the same temperature (thermal equilibrium), provided no heat is lost to the surroundings.
Formula: Heat lost by hot body = Heat gained by cold body
1.3 Important Formulae
1. Heat Energy Equation
To calculate the heat ($Q$) absorbed or released:
$$Q = m \times s \times dt$$
Where:
$m$ = mass of the body (in kg)
$s$ = specific heat capacity (in $J/kg^\circ C$)
$dt$ (or $\Delta T$) = change in temperature ($T_{final} – T_{initial}$)
2. Calorimetry Equation (Mixing)
When a hot substance (mass $m_1$, sp. heat $s_1$, temp $T_1$) is mixed with a cold substance (mass $m_2$, sp. heat $s_2$, temp $T_2$) to reach a final temperature $T$:
$$m_1 s_1 (T_1 – T) = m_2 s_2 (T – T_2)$$
1.4 Quick Review of the Chapter
Kinetic Theory of Heat
• Molecules in a substance are always in motion.
• Heat is the total sum of kinetic energy of all molecules.
• Temperature is the average kinetic energy of the molecules.
• Example: Steam causes more severe burns than boiling water because steam molecules possess higher kinetic energy (latent heat) and move rapidly, penetrating the skin.
Specific Heat Capacity
• Different substances require different amounts of heat to heat up.
• Water has a very high specific heat capacity ($4200 J/kg^\circ C$). This means it takes a lot of energy to heat water up, and it takes a long time for it to cool down.
Applications of High Specific Heat Capacity of Water
• Cooling Agent: Used in car radiators (engines) and thermal power stations because it can absorb a large amount of heat without a drastic rise in its own temperature.
• Fomentation (Hot Water Bags): Water stays hot for a longer time, making it ideal for relieving pain.
• Climate Control: Large bodies of water (seas/oceans) moderate the climate of coastal areas.
Thermal Expansion
• Generally, substances expand on heating and contract on cooling.
• Glass Tumbler: If boiling water is poured into a thick glass tumbler, it cracks. This is because the inner surface expands rapidly while the outer surface (being a poor conductor) remains cool, creating pressure that breaks the glass.
• Freezing: Water expands when it freezes into ice (Volume increases). This is why a glass bottle filled with water bursts in a deep freezer.
2.1 Choose the correct option for the following questions:
(a) Which statement defines heat?
(b) On what basis can the increase in the volume of an object be explained when it is heated?
| Order | The kinetic energy of atoms/molecules | The attraction between the atoms/molecules | Distance between the atoms/molecules |
|---|---|---|---|
| i | Increases | Decreases | Decreases |
| ii | Increases | Increase | Decreases |
| iii | Increases | Increases | Decreases |
| iv | Increases | Decreases | Increases |
(c) Specific heat capacity of a substance depends on which of the following?
(d) What is the effect of the high specific heat capacity of water?
(e) Which one of the following is the best way to insert a wide pipe into a narrower pipe?
(f) What are the lower fixed points of the thermometer in Celsius, Fahrenheit, and Kelvin scales respectively?
2.2 Differentiate between:
(i) Thermal energy and Heat
| Feature | Thermal Energy | Heat |
|---|---|---|
| Definition | The sum of the kinetic energy and potential energy of all molecules inside a body. | The form of energy that is transferred from a hot body to a cold body. |
| Nature | It is the energy stored within the system (Internal Energy). | It is energy in transit (Energy flow). |
| Dependency | Depends on the mass, temperature, and nature of the substance. | Depends on the temperature difference between two bodies. |
| Flow | Does not flow; it is the content of the body. | Always flows from a region of higher temperature to lower temperature. |
| Measurement | Cannot be measured directly (calculated theoretically). | Measured using a calorimeter (in Joules or Calories). |
(ii) Heat and Temperature
| Feature | Heat | Temperature |
|---|---|---|
| Definition | A form of energy transferred due to temperature difference. | The degree of hotness or coldness of a body (measure of average kinetic energy). |
| Cause/Effect | Heat is the cause. | Temperature is the effect. |
| Unit | SI unit is Joule (J). | SI unit is Kelvin (K); common unit is Celsius ($^\circ C$). |
| Instrument | Measured by a Calorimeter. | Measured by a Thermometer. |
| Quantity Type | Extensive property (depends on total mass). | Intensive property (independent of mass). |
(iii) The lower fixed point and the upper fixed point of a thermometer
| Feature | Lower Fixed Point (Ice Point) | Upper Fixed Point (Steam Point) |
|---|---|---|
| Definition | The temperature at which pure ice melts into water at standard pressure. | The temperature at which pure water boils into steam at standard pressure. |
| Celsius Value | $0^{\circ}C$ | $100^{\circ}C$ |
| Fahrenheit Value | $32^{\circ}F$ | $212^{\circ}F$ |
| Kelvin Value | $273 K$ | $373 K$ |
| Physical State | Represents the equilibrium between solid ice and liquid water. | Represents the equilibrium between liquid water and water vapor. |
2.3 Give reason:
(a) An iron bar heats up when it is hammered continuously for some time.
Reason: When a hammer strikes an iron bar, the mechanical energy of the hammer is transferred to the molecules of the iron. This work done against inter-atomic forces and friction increases the kinetic energy of the iron molecules. Since temperature is a measure of average kinetic energy, the increase in kinetic energy manifests as a rise in temperature, causing the bar to heat up.
(b) Tea in an open teacup stops cooling after some time.
Reason: Tea cools down by losing heat to the surroundings through evaporation and convection until it reaches the same temperature as the surrounding air (room temperature). Once the temperature of the tea equals the temperature of the surroundings, thermal equilibrium is established. At this point, the rate of heat loss equals the rate of heat gain, so the tea stops cooling further.
(c) Water pipes crack in cold places in the winter.
Reason: This occurs due to the anomalous expansion of water. When the atmospheric temperature drops below $4^{\circ}C$, water inside the pipes begins to expand as it cools down to $0^{\circ}C$ and freezes into ice. Ice has a larger volume than the same mass of liquid water. This expansion exerts a tremendous outward pressure on the inner walls of the pipe, causing them to crack or burst.
(d) To cool the car’s engine, water is kept in its radiator.
Reason: Water has a very high specific heat capacity ($4200 J/kg^{\circ}C$). This means it can absorb a large amount of heat energy from the engine with only a small rise in its own temperature. This property makes water an excellent coolant, as it effectively removes heat from the engine without boiling away quickly.
(e) The hot water bag is used to give hot pressure to the parts of a body.
Reason: Due to the high specific heat capacity of water, a hot water bag contains a large amount of heat energy for a given temperature. It releases this heat very slowly over a long period. This provides sustained warmth and relief to the affected body parts (fomentation) without cooling down immediately.
(f) There is no significant difference in temperature between the daytime and the nighttime in the coastal areas.
Reason: Water in the sea has a much higher specific heat capacity than the land. During the day, the sea absorbs solar heat slowly, keeping the air cool (Sea Breeze). At night, the sea releases the absorbed heat slowly, keeping the coastal air warm (Land Breeze). This continuous exchange of heat moderates the climate, preventing extreme temperature differences between day and night.
(g) Temperature differs a lot between the day and night in the desert.
Reason: Sand has a very low specific heat capacity. This allows it to heat up very quickly during the day when the sun is out, raising the temperature significantly. Conversely, at night, it loses that heat very rapidly, causing the temperature to drop drastically.
2.4 Answer the following questions:
(a) In the perception of a certain man, a bucket of lukewarm water contains more thermal energy than a large tank of cold water. Correct this understanding based on the definitions of thermal energy and temperature.
Answer: The man is confusing Temperature with Thermal Energy.
Temperature depends on the average kinetic energy of molecules. Lukewarm water has a higher temperature than cold water.
Thermal Energy is the total sum of kinetic energy of all molecules. Since the large tank contains a much larger mass (number of molecules) than the bucket, the total energy of all those molecules combined is likely greater than that of the fewer molecules in the bucket, even if the bucket is warmer.
(b) Touching a cup of hot tea feels hot but touching an ice cube feels cool. Explain it based on the motion of their molecules.
Answer:
Hot Tea: The molecules in hot tea have high kinetic energy. When you touch the cup, heat energy flows from the tea to your hand. Your hand’s molecules begin to vibrate faster, which your nerves perceive as a sensation of heat.
Ice Cube: The molecules in ice have very low kinetic energy. When you touch it, heat energy flows from your hand to the ice. The loss of energy from your hand slows down your skin molecules, which is perceived as a sensation of cold.
(c) If the lid of a glass bottle does not open, how may it be opened using your knowledge of the effects of heat? Explain based on the kinetic energy of molecules.
Answer: The lid can be opened by running hot water over the metal lid or dipping the neck of the bottle in warm water.
Explanation: Metal expands more than glass for the same rise in temperature. When heated, the molecules of the metal lid gain kinetic energy and move further apart (expand). This expansion increases the size of the lid slightly, loosening its grip on the glass neck, allowing it to open easily.
(d) Study the relationship between the volume and temperature of water shown in the given graph and answer the following questions.
(i) What is the special property of water shown in the graph called?
Answer: It is called the Anomalous Expansion of Water.
(ii) Mention the change in volume of water that appears when water is heated from $0^{\circ}C$ to $10^{\circ}C$.
Answer: When heated from $0^{\circ}C$ to $4^{\circ}C$, the volume of water decreases. When heated from $4^{\circ}C$ to $10^{\circ}C$, the volume of water increases. The volume is minimum at $4^{\circ}C$.
(iii) How does the property of water shown in the graph differ from the property of other liquids?
Answer: Most other liquids expand uniformly (volume increases) when heated from solid to liquid and as temperature rises. Water is unique because it contracts when heated from $0^{\circ}C$ to $4^{\circ}C$.
(iv) Explain the importance of this knowledge in daily life.
Answer: This property allows aquatic life to survive in frozen lakes. Since water is densest at $4^{\circ}C$, it sinks to the bottom, while ice (at $0^{\circ}C$) floats on top. This ice layer insulates the water below, keeping it at a survivable $4^{\circ}C$ for fish and other organisms during winter.
(v) Draw a model graph to show the relationship between the density of water and temperature.
(e) Once in winter, while drinking the water from a steel jug on the table, Samir felt the water to be warmer towards the bottom. Justify his experience based on scientific facts.
Answer: This is due to the anomalous expansion of water. In cold winter conditions, as water cools, the water at $4^{\circ}C$ becomes the densest and sinks to the bottom of the jug. The water closer to the surface may be cooler (between $0^{\circ}C$ and $4^{\circ}C$) because it is less dense. Therefore, the bottom layer remains at $4^{\circ}C$ while the top layer is colder, making the bottom feel relatively warmer.
(f) What are the differences between the process of freezing ghee and honey in terms of their volume and density?
Answer:
Ghee: Like most substances, when ghee freezes (solidifies), its molecules pack tighter. Therefore, its volume decreases and its density increases.
Honey: Honey contains water. When it freezes (crystallizes), the behavior is influenced by the water content and sugar crystals. Generally, the water component expands. Thus, freezing honey often leads to a slight increase in volume and a decrease in density relative to the pure liquid phase, similar to water freezing into ice.
(h) Of the two ice cubes of identical shape, one is kept in an aluminum box and the other in a wooden box. Which ice cube melts faster? Explain in terms of the melting process.
Answer: The ice cube in the aluminum box will melt faster.
Reason: Aluminum is a good conductor of heat, while wood is an insulator. The aluminum box will rapidly conduct heat from the surroundings to the ice, accelerating the melting process. The wooden box blocks heat transfer, keeping the ice frozen for longer.
(i) What is specific heat capacity? Write its SI unit.
Answer: Specific heat capacity is the amount of heat energy required to raise the temperature of 1 kg of a substance by $1^{\circ}C$ (or 1 Kelvin).
SI Unit: Joule per kilogram per degree Celsius ($J/kg^{\circ}C$) or Joule per kilogram Kelvin ($J/kg \cdot K$).
(j) What is the heat equation?
Answer: The heat equation relates heat energy ($Q$), mass ($m$), specific heat capacity ($s$), and change in temperature ($dt$ or $\Delta T$).
Formula: $Q = m \times s \times dt$
(k) Write any two applications of specific heat capacity.
Answer:
• Car Radiators: Water is used as a coolant because its high specific heat capacity allows it to absorb significant engine heat.
• Fomentation (Hot Water Bags): Used for relieving pain because water retains heat for a long time due to high specific heat capacity.
(l) Describe the condition in which water can be boiled at a temperature less than $100^{\circ}C$.
Answer: Water boils when its vapor pressure equals the atmospheric pressure. At sea level, this happens at $100^{\circ}C$. However, if the atmospheric pressure is reduced (e.g., at high altitudes or by using a vacuum pump), the boiling point decreases. Thus, water can be boiled at less than $100^{\circ}C$ in low-pressure environments.
(m) Write the types of thermometers used in daily life. Also, mention their working principles.
Answer:
• Clinical Thermometer: Used to measure human body temperature. Working principle: Thermal expansion of mercury (or alcohol) in a constricted capillary tube.
• Laboratory Thermometer: Used for science experiments. Working principle: Uniform thermal expansion of liquid (mercury/alcohol) with temperature.
• Digital Thermometer: Used for quick reading. Working principle: Uses a thermistor (heat-sensitive resistor) whose electrical resistance changes with temperature.
(n) What is thermometer calibration? Describe the method.
Answer: Calibration is the process of marking the temperature scale on a thermometer.
Method:
1. Lower Fixed Point: The thermometer bulb is placed in pure melting ice. The level of mercury is marked as $0^{\circ}C$.
2. Upper Fixed Point: The thermometer is placed in steam from boiling pure water. The level is marked as $100^{\circ}C$.
3. Division: The distance between the $0^{\circ}C$ and $100^{\circ}C$ marks is divided into 100 equal parts. Each part represents $1^{\circ}C$.
2.5 Solve the following mathematical problems:
(a) Calculate the amount of heat required to raise the temperature of 500 g of water from $15^{\circ}C$ to $85^{\circ}C$.
Given:
Mass ($m$) = $500 \text{ g} = 0.5 \text{ kg}$
Initial Temp ($t_1$) = $15^{\circ}C$
Final Temp ($t_2$) = $85^{\circ}C$
Change in Temp ($dt$) = $85 – 15 = 70^{\circ}C$
Specific heat of water ($s$) = $4200 J/kg^{\circ}C$ (standard value)
Formula: $Q = m \times s \times dt$
Calculation:
$Q = 0.5 \times 4200 \times 70$
$Q = 2100 \times 70$
$Q = 147,000 J$
$Q = 147 kJ$
Answer: The heat required is 147 kJ.
(b) The specific heat capacity of iron is $460J/kg^{\circ}C$. Calculate the heat released by an iron sphere of mass 5kg while cooling it from $430^{\circ}C$ to $30^{\circ}C$.
Given:
Mass ($m$) = $5 \text{ kg}$
Specific heat ($s$) = $460 J/kg^{\circ}C$
Initial Temp ($t_1$) = $430^{\circ}C$
Final Temp ($t_2$) = $30^{\circ}C$
Change in Temp ($dt$) = $430 – 30 = 400^{\circ}C$
Formula: $Q = m \times s \times dt$
Calculation:
$Q = 5 \times 460 \times 400$
$Q = 2300 \times 400$
$Q = 920,000 J$
$Q = 920 kJ$
Answer: The heat released is 920 kJ.
Note: The textbook answer says 920J. This is a clear misprint; the correct mathematical result is 920,000J or 920kJ.
(c) If 2kg of paraffin needs 4200J of heat to increase its temperature through $10^{\circ}C$, calculate the amount of heat required to increase the temperature of 4kg of paraffin from $20^{\circ}C$ to $40^{\circ}C$.
Step 1: Find Specific Heat ($s$) of paraffin from the first case.
Mass ($m_1$) = $2 \text{ kg}$
Heat ($Q_1$) = $4200 \text{ J}$
Change in Temp ($dt_1$) = $10^{\circ}C$
$s = Q_1 / (m_1 \times dt_1)$
$s = 4200 / (2 \times 10) = 4200 / 20 = 210 J/kg^{\circ}C$
Step 2: Calculate Heat ($Q_2$) for the second case.
Mass ($m_2$) = $4 \text{ kg}$
Change in Temp ($dt_2$) = $40 – 20 = 20^{\circ}C$
Specific Heat ($s$) = $210 J/kg^{\circ}C$ (calculated above)
$Q_2 = m_2 \times s \times dt_2$
$Q_2 = 4 \times 210 \times 20$
$Q_2 = 840 \times 20$
$Q_2 = 16,800 J$
Answer: The heat required is 16,800 J.
Note: The provided book answer is $8.4 \times 10^4 J$ (84,000 J). This discrepancy suggests the book numbers for the first part might contain a typo—perhaps the first heat was meant to be 21,000J, or the specific heat of paraffin was assumed to be higher initially. Based strictly on the provided numbers, 16,800 J is the correct calculated answer.
(d) If a substance of mass 500g needs 7938J of heat to increase its temperature from $100^{\circ}C$ to $226^{\circ}C$, calculate its specific heat capacity.
Given:
Mass ($m$) = $500 \text{ g} = 0.5 \text{ kg}$
Heat ($Q$) = $7938 \text{ J}$
Change in Temp ($dt$) = $226 – 100 = 126^{\circ}C$
Formula: $s = Q / (m \times dt)$
Calculation:
$s = 7938 / (0.5 \times 126)$
$s = 7938 / 63$
$s = 126 J/kg^{\circ}C$
Answer: The specific heat capacity is $126 J/kg^{\circ}C$.
(e) A bucket contains 16kg of water at $25^{\circ}C$. Calculate the temperature of the mixture formed when 4 kg of water at $80^{\circ}C$ is mixed with it. (Here, the heat lost to the surrounding is neglected)
Given:
Cold Water: Mass ($m_1$) = $16 \text{ kg}$, Temp ($t_1$) = $25^{\circ}C$
Hot Water: Mass ($m_2$) = $4 \text{ kg}$, Temp ($t_2$) = $80^{\circ}C$
Specific heat ($s$) is the same for both (cancels out).
Let final temperature be $T$.
Principle: Heat Lost by Hot Body = Heat Gained by Cold Body
$m_2 \times s \times (t_2 – T) = m_1 \times s \times (T – t_1)$
$4 \times (80 – T) = 16 \times (T – 25)$
Calculation:
Divide both sides by 4:
$(80 – T) = 4 \times (T – 25)$
$80 – T = 4T – 100$
$80 + 100 = 4T + T$
$180 = 5T$
$T = 180 / 5$
$T = 36^{\circ}C$
Answer: The final temperature of the mixture is $36^{\circ}C$.