Frank Brothers Solutions for ICSE Class 10 Physics Chapter 5.2 Heat Calorimetry

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Question 1. What is thermal energy? Give an example.
Answer: Thermal energy refers to the energy that comes from a heat source. For instance, an electric room heater produces thermal energy to heat up a chilly indoor space during winter.
In simple words: Thermal energy is heat energy. An example is the warmth we get from a room heater.

Exam Tip: Always provide a clear everyday example when defining types of energy to secure full marks.

Question 2. Is heat a form of energy?
Answer: Indeed, heat is classified as a type of energy.
In simple words: Yes, heat is definitely a kind of energy.

Exam Tip: This is a direct conceptual question; a simple "Yes" followed by a brief statement is sufficient.

Question 3. Define temperature.
Answer: Temperature is a physical characteristic that measures our everyday feeling of hot and cold using numbers. It represents how hot or cold an object or its surroundings are.
In simple words: Temperature is a measurement that tells us how hot or cold something is.

Exam Tip: Remember to use keywords like "degree of hotness" or "measurement of coldness" in your definition.

Question 4. Differentiate between heat and temperature.
Answer: The differences between heat and temperature are summarized in the table below:

In simple words: Heat is the energy itself, while temperature is just the measurement of how hot or cold something gets because of that heat.

Exam Tip: Highlighting that heat is the cause and temperature is the effect is a key point often looked for by examiners.

Question 5. What is the SI unit of heat energy?
Answer: The standard SI unit for measuring thermal energy is the joule, symbolized by J.
In simple words: The standard scientific unit of heat is the joule (J).

Exam Tip: Always write both the name of the unit (joule) and its symbol (J) in your answer.

Question 6. Define the unit '1 joule' of heat energy.
Answer: One joule is defined as the quantity of heat needed to increase the temperature of a \( 1\text{ kg} \) mass of a substance by \( 1^\circ\text{C} \), assuming the substance has a specific heat capacity of \( 1\text{ J kg}^{-1}\text{ K}^{-1} \).
In simple words: One joule is the heat energy needed to raise the temperature of a 1 kg block of a basic material by 1 degree.

Exam Tip: Make sure to include all three essential values (1 kg, 1 degree, and specific heat capacity of \( 1\text{ J kg}^{-1}\text{ K}^{-1} \)) in your definition.

Question 7. State the relationship between a calorie and a joule. Which of these is a larger unit of energy?
Answer: The mathematical relationship is \( 1\text{ calorie (cal)} \approx 4.2\text{ joules (J)} \). Consequently, a calorie is a larger unit of energy than a joule.
In simple words: One calorie is equal to about 4.2 joules, which means a calorie is the larger unit.

Exam Tip: Remember the conversion factor \( 1\text{ cal} \approx 4.2\text{ J} \) as it is crucial for solving numerical problems.

Question 8. Name the device used to measure temperature.
Answer: Temperature is measured using an instrument called a thermometer.
In simple words: We use a thermometer to check how hot or cold something is.

Exam Tip: This is a straightforward, one-mark question. Be sure to spell "thermometer" correctly.

Question 9. Plot graphs showing the conversion between the Celsius and Fahrenheit temperature scales: (i) Celsius temperature against Fahrenheit temperature, and (ii) Fahrenheit temperature against Celsius temperature.
Answer: The conversion graphs are plotted below:
(i) Graph of Celsius temperature (\( ^\circ\text{C} \)) against Fahrenheit temperature (\( ^\circ\text{F} \)):

(ii) Graph of Fahrenheit temperature (\( ^\circ\text{F} \)) against Celsius temperature (\( ^\circ\text{C} \)):

In simple words: These graphs are straight lines because the temperature conversion formulas are linear relations.

Exam Tip: Remember that both graphs are straight lines, where \( 0^\circ\text{C} \) matches with \( 32^\circ\text{F} \) and \( 100^\circ\text{C} \) matches with \( 212^\circ\text{F} \).

Question 10. What physical quantity measures the degree of hotness of a body?
Answer: The physical property that quantifies how hot an object is is called temperature.
In simple words: Temperature is what we use to measure how hot something is.

Exam Tip: This is a basic definition. Remember to link "degree of hotness" directly with "temperature".

Question 11. What happens to the internal energy of a substance when it is heated?
Answer: When a substance is heated, its internal thermal energy rises.
In simple words: Heating something up makes its molecules gain more energy.

Exam Tip: On heating, the kinetic energy of the molecules increases, which increases the total internal energy of the substance.

Question 12. Why do gas molecules move about freely?
Answer: Molecules in a gas have extremely weak or negligible intermolecular bonds, and the distances separating them are very large. Consequently, gas molecules can travel freely in all directions.
In simple words: Gas molecules move freely because they are far apart and not held together by strong forces.

Exam Tip: Mention both the negligible intermolecular forces of attraction and the large intermolecular spaces to write a complete answer.

Question 13. Name two commonly used scales for measuring temperature.
Answer: The two widely used temperature measurement scales are:
1. The Celsius scale
2. The Fahrenheit scale
In simple words: We usually measure temperature using either the Celsius scale or the Fahrenheit scale.

Exam Tip: Although Celsius and Fahrenheit are common, Kelvin is the SI unit of temperature. Keep this distinction in mind.

Question 14. Which type of thermometer is most commonly used in everyday life?
Answer: A "liquid-in-glass" type of thermometer is the most frequently used instrument for daily temperature checks.
In simple words: The most common thermometer is the one where liquid rises inside a glass tube.

Exam Tip: The liquids typically used in these thermometers are mercury or alcohol.

Question 15. What is a doctor's thermometer also known as?
Answer: A doctor's thermometer is alternatively referred to as a clinical thermometer.
In simple words: The thermometer a doctor uses to check your fever is called a clinical thermometer.

Exam Tip: Clinical thermometers have a kink (constriction) in their capillary tube to prevent the mercury from falling back instantly.

Question 16. What is the melting point of ice on both the Celsius and Fahrenheit scales?
Answer: The melting point of ice on the two scales is as follows:
- On the Celsius scale: \( 0^\circ\text{C} \)
- On the Fahrenheit scale: Using the formula \( F = \frac{9}{5}C + 32 \):
\[ F = \frac{9}{5}(0) + 32 = 32^\circ\text{F} \]
In simple words: Ice melts at \( 0^\circ\text{C} \), which is exactly the same temperature as \( 32^\circ\text{F} \).

Exam Tip: Keep the conversion formula \( F = \frac{9}{5}C + 32 \) memorized to convert any temperature scale quickly.

Question 17. Why are the Celsius and Fahrenheit scales commonly used for measuring temperature? State their reference points?
Answer: The Celsius and Fahrenheit scales are widely adopted because they are anchored on the properties of water. The Celsius scale sets the freezing point of water at \( 0^\circ\text{C} \) and its boiling point at \( 100^\circ\text{C} \). On the Fahrenheit scale, these same reference temperatures are marked at \( 32^\circ\text{F} \) and \( 212^\circ\text{F} \) respectively.
In simple words: These scales are popular because they use the freezing and boiling points of water as their main markers.

Exam Tip: Know the lower and upper fixed points of both scales as they define the standard interval used in calibration.

Question 18. What is the normal body temperature of a healthy human being on the Celsius scale?
Answer: A healthy human body has a standard temperature of \( 37^\circ\text{C} \).
In simple words: A healthy person has a body temperature of 37 degrees Celsius.

Exam Tip: In Fahrenheit, this corresponds to \( 98.6^\circ\text{F} \). Try to remember both values.

Question 19. Write the formula to convert temperature from Fahrenheit scale to Celsius scale.
Answer: The formula to convert Fahrenheit (\( F \)) into Celsius (\( C \)) is given by:
\[ C = \frac{5}{9}(F - 32) \]
In simple words: Subtract 32 from the Fahrenheit temperature and multiply by 5/9 to get the temperature in Celsius.

Exam Tip: Always perform the subtraction inside the parentheses first before multiplying by \( \frac{5}{9} \) to avoid calculation errors.

Question 20. What are the lower and upper fixed points on the Fahrenheit temperature scale?
Answer: On the Fahrenheit scale, the reference points are:
- Lower fixed point: \( 32^\circ\text{F} \)
- Upper fixed point: \( 212^\circ\text{F} \)
In simple words: On the Fahrenheit scale, the bottom point is 32 and the top point is 212.

Exam Tip: These points represent the melting point of ice and the boiling point of pure water under standard atmospheric pressure.

Question 21. Explain how the Celsius scale is defined and calibrated using reference points.
Answer: On the Celsius scale, the melting point of ice is defined as the "lower fixed point" (\( 0^\circ\text{C} \)) and the boiling point of pure water is defined as the "upper fixed point" (\( 100^\circ\text{C} \)). The interval between these two reference markers is divided into exactly 100 equal parts. Each of these equal divisions represents a temperature change of "one degree Celsius" (\( 1^\circ\text{C} \)).
In simple words: The Celsius scale starts at 0 for melting ice and ends at 100 for boiling water, with 100 equal steps in between.

Exam Tip: Explain both fixed points and clearly state how the 100 divisions are derived to get full marks.

Question 22. Which represents a greater temperature rise: \( 20^\circ\text{C} \) or \( 20^\circ\text{F} \)? Show with calculations.
Answer: To compare the two values, we convert \( 20^\circ\text{C} \) to the Fahrenheit scale using the formula:
\[ F = \frac{9}{5}C + 32 \]
Substituting \( C = 20 \):
\[ F = \frac{9}{5}(20) + 32 = 36 + 32 = 68^\circ\text{F} \]
This means that \( 20^\circ\text{C} \) is equivalent to \( 68^\circ\text{F} \). Since \( 68^\circ\text{F} > 20^\circ\text{F} \), we can write:
\[ 20^\circ\text{C} > 20^\circ\text{F} \]
Hence, a temperature rise of \( 20^\circ\text{C} \) represents a significantly larger temperature change than \( 20^\circ\text{F} \).
In simple words: 20 degrees Celsius is equal to 68 degrees Fahrenheit, which is much warmer than 20 degrees Fahrenheit.

Exam Tip: Always perform the step-by-step conversion using the standard conversion formula to substantiate your answer in conceptual numericals.

Question 23. At what temperature do the Celsius and Fahrenheit scales show the same reading? Prove mathematically.
Answer: Let the temperature at which both scales show the same value be \( x \). Using the standard conversion relation:
\[ F = \frac{9}{5}C + 32 \]
We set both \( F \) and \( C \) to \( x \):
\[ x = \frac{9}{5}x + 32 \]
Multiplying the entire equation by 5:
\[ 5x = 9x + 160 \]
Rearranging the terms:
\[ -4x = 160 \]
\[ x = -40 \]
Therefore, at \( -40^\circ \), the readings on both temperature scales are completely identical, meaning:
\[ -40^\circ\text{C} \equiv -40^\circ\text{F} \]
In simple words: At minus 40 degrees, the Celsius and Fahrenheit scales show exactly the same number.

Exam Tip: This is a classic exam question. Proving it by setting \( F = C = x \) is the most elegant way to solve it.

Question 24. Convert \( 212^\circ\text{F} \) into the Celsius scale.
Answer: We convert \( 212^\circ\text{F} \) to the Celsius scale using the formula:
\[ C = \frac{5}{9}(F - 32) \]
Substituting \( F = 212 \):
\[ C = \frac{5}{9}(212 - 32) \]
\[ C = \frac{5}{9}(180) = 5 \times 20 = 100^\circ\text{C} \]
Thus, \( 212^\circ\text{F} \) is equal to \( 100^\circ\text{C} \).
In simple words: 212 degrees Fahrenheit is exactly equal to 100 degrees Celsius, which is the boiling point of water.

Exam Tip: Always double-check your arithmetic by remembering that these two temperatures represent the boiling point of pure water on their respective scales.

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Question 25. Express absolute zero (\( 0\text{ K} \)) on: (a) the Celsius scale, and (b) the Fahrenheit scale.
Answer:
(a) To convert \( 0\text{ K} \) to the Celsius scale, we use the relationship:
\[ C = K - 273 \]
Substituting \( K = 0 \):
\[ C = 0 - 273 = -273^\circ\text{C} \]
Thus, \( 0\text{ K} \) is equivalent to \( -273^\circ\text{C} \).

(b) To convert this temperature to the Fahrenheit scale, we substitute \( C = -273 \) into the Fahrenheit formula:
\[ F = \frac{9}{5}C + 32 \]
\[ F = \frac{9}{5}(-273) + 32 \]
\[ F = -491.4 + 32 = -459.4^\circ\text{F} \]
Thus, \( 0\text{ K} \) is equivalent to \( -459.4^\circ\text{F} \).
In simple words: Absolute zero, which is 0 Kelvin, is equal to minus 273 degrees Celsius or minus 459.4 degrees Fahrenheit.

Exam Tip: Note that \( 273.15 \) is the precise conversion value, but using \( 273 \) is generally accepted in high school exams.

Question 26. Define absolute zero. What are its values on different temperature scales?
Answer: Absolute zero is defined as the theoretical temperature at which the volume or pressure of an ideal gas drops to zero. It represents the lowest possible temperature where molecular motion ceases. Its values are \( 0\text{ K} \) on the Kelvin scale, which is equal to \( -273^\circ\text{C} \) or \( -459.4^\circ\text{F} \).
In simple words: Absolute zero is the coldest possible temperature where gas molecules stop moving completely.

Exam Tip: Mention the zero volume/pressure of an ideal gas to write a scientifically complete definition.

Question 27. If a body has a temperature of \( 20^\circ\text{C} \), what is its corresponding temperature on the Kelvin scale?
Answer: We convert Celsius (\( C \)) to Kelvin (\( K \)) using the formula:
\[ K = C + 273 \]
Substituting \( C = 20 \):
\[ K = 20 + 273 = 293\text{ K} \]
Thus, the temperature of the body on the Kelvin scale is \( 293\text{ K} \).
In simple words: To convert Celsius to Kelvin, simply add 273 to the Celsius temperature.

Exam Tip: Never write a degree symbol (\( ^\circ \)) with the Kelvin unit; write it simply as "K".

Question 28. Convert a normal human body temperature of \( 37^\circ\text{C} \) into the Fahrenheit scale.
Answer: To find the equivalent temperature in Fahrenheit, we use the conversion formula:
\[ F = \frac{9}{5}C + 32 \]
Substituting \( C = 37 \):
\[ F = \frac{9}{5}(37) + 32 \]
\[ F = 66.6 + 32 = 98.6^\circ\text{F} \]
Therefore, \( 37^\circ\text{C} \) is equal to \( 98.6^\circ\text{F} \).
In simple words: 37 degrees Celsius is equal to 98.6 degrees Fahrenheit, which is normal human body temperature.

Exam Tip: Practice conversions with fractional numbers to ensure you perform the multiplication step precisely during exams.

Question 29. State the SI units of: (i) Amount of heat, (ii) Heat capacity, and (iii) Specific heat capacity.
Answer: The SI units for these physical quantities are:
(i) Amount of heat: \( \text{joule (J)} \)
(ii) Heat capacity: \( \text{joule per Kelvin (J K}^{-1}\text{)} \)
(iii) Specific heat capacity: \( \text{joule per kilogram per Kelvin (J kg}^{-1}\text{ K}^{-1}\text{)} \)
In simple words: We measure heat in Joules, heat capacity in Joules per Kelvin, and specific heat capacity in Joules per kilogram per Kelvin.

Exam Tip: Always write both the full name and the symbolic representation of units to avoid losing marks.

Question 30. A mass of 2 kg of water at \( 80^\circ\text{C} \) is mixed with 8 kg of water at \( 25^\circ\text{C} \). Calculate the final temperature of the mixture, neglecting any heat loss to the surroundings.
Answer: Let \( C \) be the specific heat capacity of water, and let \( \theta \) represent the final equilibrium temperature of the mixture.
Using the principle of calorimetry:
\[ \text{Heat lost by hot water} = \text{Heat gained by cold water} \]
\[ m_1 \times C \times (T_1 - \theta) = m_2 \times C \times (\theta - T_2) \]
Substituting the given values:
\[ 2 \times C \times (80 - \theta) = 8 \times C \times (\theta - 25) \]
Dividing both sides by the specific heat capacity \( C \):
\[ 2(80 - \theta) = 8(\theta - 25) \]
Dividing both sides by 2:
\[ 80 - \theta = 4(\theta - 25) \]
\[ 80 - \theta = 4\theta - 100 \]
\[ 5\theta = 180 \]
\[ \theta = 36 \]
Therefore, the final temperature of the water mixture will be \( 36^\circ\text{C} \).
In simple words: When 2 kg of hot water at 80°C is mixed with 8 kg of cold water at 25°C, the final temperature settles at 36°C.

Exam Tip: Always cancel out common factors like specific heat capacity \( C \) early in your equation to make calculations simpler.

Question 31. A liquid A of mass \( m \) and specific heat capacity \( 0.84\text{ J g}^{-1}\text{ K}^{-1} \) at \( 40^\circ\text{C} \) is mixed with \( 100\text{ g} \) of liquid B of specific heat capacity \( 2.1\text{ J g}^{-1}\text{ K}^{-1} \) at \( 20^\circ\text{C} \). If the final temperature of the mixture is \( 32^\circ\text{C} \), find the mass \( m \) of liquid A, assuming no loss of heat.
Answer: Let \( m \) be the mass of liquid A.
Assuming that there is no heat loss to the surroundings, we can use the principle of calorimetry:
\[ \text{Heat energy lost by liquid A} = \text{Heat energy gained by liquid B} \]
\[ m \times C_A \times (T_A - \theta) = m_B \times C_B \times (\theta - T_B) \]
Substituting the given values:
\[ m \times 0.84 \times (40 - 32) = 100 \times 2.1 \times (32 - 20) \]
\[ m \times 0.84 \times 8 = 100 \times 2.1 \times 12 \]
\[ m \times 6.72 = 2520 \]
\[ m = \frac{100 \times 2.1 \times 12}{0.84 \times 8} = 375\text{ g} \]
Thus, the mass of liquid A is \( 375\text{ g} \).
In simple words: By equating the heat lost by the warmer liquid to the heat gained by the cooler one, we find the mass of liquid A is 375 grams.

Exam Tip: Ensure that the units of specific heat capacity and mass are consistent (e.g., both in grams and Joules) before executing calculations.

Question 32. What is the specific heat capacity of water in SI units?
Answer: The specific heat capacity of water is \( 4200\text{ J kg}^{-1}\text{ K}^{-1} \).
In simple words: Water has a specific heat capacity of 4200 Joules per kilogram per Kelvin.

Exam Tip: This is an important standard physical constant that you should memorize, as it is often not provided in numeric questions.

Question 33. What is meant by the statement 'the specific heat capacity of water is \( 4200\text{ J kg}^{-1}\text{ K}^{-1} \)'?
Answer: This statement signifies that \( 4200\text{ J} \) of heat energy is needed to raise the temperature of \( 1\text{ kg} \) of water by a temperature difference of \( 1\text{ K} \) (or \( 1^\circ\text{C} \)).
In simple words: It means you need 4200 Joules of heat to warm up 1 kilogram of water by 1 degree.

Exam Tip: Be sure to mention both "1 kg" and "1 K" in your explanation to define the concept completely.

Question 34. State two everyday applications of the high specific heat capacity of water.
Answer: Two major applications of water's high specific heat capacity are:
(i) As a cooling agent: Due to its ability to absorb large amounts of heat with a small temperature rise, water is used as a coolant in automobile radiators and industrial engines.
(ii) As a heat reservoir: In very cold countries, water is used as a heat reservoir around wine and juice bottles to prevent them from freezing. Because water has a high specific heat capacity, it can supply a large amount of heat as it cools, keeping the surrounding beverages above their freezing points.
In simple words: Water is used to cool engines because it absorbs a lot of heat, and it is used to prevent food from freezing because it stays warm for a long time.

Exam Tip: Use scientific terms like "coolant" and "heat reservoir" and explain them using the high specific heat capacity of water.

Question 35. What is a calorimeter? Why is it made of copper?
Answer: A calorimeter is an instrument employed to measure the amount of heat energy exchanged during a physical or chemical process. It is typically constructed from copper for two main reasons:
1. High thermal conductivity: Copper is an excellent conductor of heat, allowing the calorimeter to rapidly reach thermal equilibrium with its contents.
2. Low specific heat capacity: With a low specific heat capacity of \( 390\text{ J kg}^{-1}\text{ K}^{-1} \), the copper container absorbs a negligible fraction of the heat energy released during the experiment, thereby minimizing experimental errors.
In simple words: A calorimeter measures heat. It is made of copper because copper conducts heat quickly and absorbs very little of the heat being measured.

Exam Tip: Focus on the two key properties of copper: "high conductivity" and "low specific heat capacity" to score maximum marks.

Question 36. A copper vessel has a heat capacity of \( 966\text{ J }^\circ\text{C}^{-1} \). (i) Find the heat energy required to raise its temperature by \( 15^\circ\text{C} \). (ii) If the mass of the vessel is \( 2\text{ kg} \), calculate its specific heat capacity.
Answer:
(i) We are given the heat capacity of the vessel (\( C' = m \times c \)) as \( 966\text{ J }^\circ\text{C}^{-1} \).
The heat energy required (\( Q \)) is calculated using the formula:
\[ Q = C' \times \Delta T \]
\[ Q = 966 \times 15 = 14490\text{ J} \]

(ii) The relationship between heat capacity (\( C' \)) and specific heat capacity (\( c \)) is:
\[ C' = m \times c \]
Rearranging for specific heat capacity (\( c \)):
\[ c = \frac{C'}{m} \]
Given mass \( m = 2\text{ kg} \):
\[ c = \frac{966}{2} = 483\text{ J kg}^{-1}\text{ }^\circ\text{C}^{-1} \]
In simple words: We multiply heat capacity by the temperature change to find the heat energy (14,490 Joules), and divide by mass to find the specific heat capacity (483 J/kg°C).

Exam Tip: Note that specific heat capacity can be expressed in either \( \text{J kg}^{-1}\text{ }^\circ\text{C}^{-1} \) or \( \text{J kg}^{-1}\text{ K}^{-1} \) as a temperature interval of \( 1^\circ\text{C} \) is equal to \( 1\text{ K} \).

Question 37. Why do farmers fill their agricultural fields with water on cold winter nights?
Answer: Farmers irrigate their fields on freezing winter nights to prevent crops from being damaged by frost. If there is no water, and the environmental temperature falls below \( 0^\circ\text{C} \), the water inside the plant capillaries (veins) will freeze. Due to the anomalous expansion of water, freezing ice expands and occupies more volume than liquid water, causing the delicate capillaries to rupture and destroy the crops. Sprinkling water provides a thermal buffer; because of water's high specific heat capacity and high latent heat of fusion, it releases large quantities of heat upon cooling, preventing the temperature of the plants from falling below \( 0^\circ\text{C} \).
In simple words: Farmers put water on crops because freezing water expands and can burst the plants' veins. The water keeps the plants from getting too cold.

Exam Tip: Always mention "anomalous expansion of water" and "bursting of plant veins" as they are the key scientific concepts behind this phenomenon.

Question 38. A mass of 300 g of hot water at \( 50^\circ\text{C} \) is mixed with 600 g of cold water. The temperature of the cold water rises by \( 15^\circ\text{C} \) during the process. If there is no heat loss to the surroundings, calculate: (i) the final temperature of the mixture, and (ii) the initial temperature of the cold water.
Answer: Let the initial temperature of the cold water be \( t \), and the final equilibrium temperature of the mixture be \( \theta \).
The rise in temperature of the cold water is \( \theta - t = 15^\circ\text{C} \).
Using the principle of calorimetry:
\[ \text{Heat gained by cold water} = \text{Heat lost by hot water} \]
\[ 600 \times C \times 15 = 300 \times C \times (50 - \theta) \]
Dividing both sides by \( 300 \times C \):
\[ 2 \times 15 = 50 - \theta \]
\[ 30 = 50 - \theta \]
\[ \theta = 20^\circ\text{C} \]
Thus, the final temperature of the mixture is \( 20^\circ\text{C} \).

To find the initial temperature \( t \):
\[ \theta - t = 15^\circ\text{C} \]
\[ 20 - t = 15 \]
\[ t = 5^\circ\text{C} \]
Therefore, the initial temperature of the cold water was \( 5^\circ\text{C} \).
In simple words: The final temperature of the mixture is 20°C, and the cold water started at an initial temperature of 5°C.

Exam Tip: Keep your equations structured by dividing out the common specific heat capacity factor \( C \) first to make calculations straightforward.

Question 39. Define heat capacity of a body. State its dependency factors and units.
Answer: The heat capacity of a body is defined as the amount of heat energy required to raise its temperature by \( 1^\circ\text{C} \) (or \( 1\text{ K} \)). It is dependent on both the total mass of the body and its material composition.
Units: \( \text{J }^\circ\text{C}^{-1} \) or \( \text{cal }^\circ\text{C}^{-1} \).
In simple words: Heat capacity is how much heat a whole object needs to warm up by 1 degree. It depends on how heavy it is and what it is made of.

Exam Tip: Do not confuse heat capacity (dependent on mass) with specific heat capacity (independent of mass).

Question 40. Define specific heat capacity. State its standard SI and cgs units.
Answer: Specific heat capacity is the amount of heat energy required to increase the temperature of a unit mass (\( 1\text{ kg} \) or \( 1\text{ g} \)) of a substance by \( 1^\circ\text{C} \) (or \( 1\text{ K} \)).
Units: The SI unit is \( \text{J kg}^{-1}\text{ K}^{-1} \), and the CGS unit is \( \text{cal g}^{-1}\text{ }^\circ\text{C}^{-1} \).
In simple words: Specific heat capacity is the heat needed to warm up exactly 1 kg of a material by 1 degree.

Exam Tip: Remember that specific heat capacity is a characteristic property of a substance and does not depend on its mass.

Question 41. A mass of \( 0.5\text{ kg} \) of lemon squash at \( 30^\circ\text{C} \) is placed in a refrigerator to cool it down to \( 5^\circ\text{C} \). If heat is extracted by the refrigerator at a constant rate of \( 30\text{ J s}^{-1} \), calculate the time required for this cooling process. (Take the specific heat capacity of lemon squash as \( 4200\text{ J kg}^{-1}\text{ K}^{-1} \)).
Answer: First, calculate the total change in temperature of the lemon squash:
\[ \Delta T = 30 - 5 = 25^\circ\text{C} \]
Now, find the total heat energy (\( Q \)) that must be extracted:
\[ Q = m \times C \times \Delta T \]
\[ Q = 0.5 \times 4200 \times 25 = 52500\text{ J} \]

Let \( t \) be the time taken in seconds. The constant rate of heat extraction (\( P \)) is given as \( 30\text{ J s}^{-1} \):
\[ P = \frac{Q}{t} \]
\[ 30 = \frac{52500}{t} \]
\[ t = \frac{52500}{30} = 1750\text{ s} \]
Converting seconds to minutes:
\[ t = \frac{1750}{60} \approx 29.2\text{ min} \]
Therefore, the time required to cool the lemon squash is \( 29.2\text{ minutes} \).
In simple words: We find that 52,500 Joules of heat must be removed. At a rate of 30 Joules per second, it takes 1750 seconds, which is about 29.2 minutes.

Exam Tip: Keep track of your units. Remember that rate is in Joules per second (\( \text{J s}^{-1} \)), so the calculated time \( t \) is initially in seconds and must be converted to minutes.

Question 42. Describe the experimental method to determine the specific heat capacity of a solid using the method of mixtures (calorimeter), and derive the mathematical expression for it.
Answer: To find the specific heat capacity of a solid, follow these experimental steps:
1. Weigh the given solid using a balance.
2. Heat the solid by placing it inside a beaker of boiling water until it reaches a constant, high temperature.
3. Weigh the empty calorimeter along with its stirrer.
4. Fill the calorimeter with cold water and weigh it again to find the mass of the water used.
5. Record the initial temperature of the water and the calorimeter.
6. Quickly transfer the hot solid into the calorimeter. Stir the contents gently and record the final steady temperature of the mixture.

Let:
- Mass of calorimeter with stirrer = \( m_1\text{ g} \)
- Specific heat capacity of calorimeter = \( C_1 \)
- Mass of water taken = \( m_2\text{ g} \)
- Specific heat capacity of water = \( C_2 \)
- Mass of solid = \( m_3\text{ g} \)
- Specific heat capacity of the solid = \( C_3 \)
- Initial temperature of the solid = \( x^\circ\text{C} \)
- Initial temperature of water and calorimeter = \( y^\circ\text{C} \)
- Final equilibrium temperature of the mixture = \( z^\circ\text{C} \)

Applying the principle of calorimetry (assuming no heat is lost to the surroundings):
\[ \text{Heat lost by solid} = \text{Heat gained by calorimeter} + \text{Heat gained by water} \]
\[ m_3 C_3 (x - z) = m_1 C_1 (z - y) + m_2 C_2 (z - y) \]
\[ m_3 C_3 (x - z) = (m_1 C_1 + m_2 C_2)(z - y) \]
\[ C_3 = \frac{(m_1 C_1 + m_2 C_2)(z - y)}{m_3 (x - z)} \]
This formula helps us calculate the specific heat capacity of the solid.
In simple words: We measure the masses and initial temperatures of the solid, water, and calorimeter, mix them, record the final temperature, and use the calorimetry formula to calculate the solid's specific heat capacity.

Exam Tip: Ensure you label each term clearly in the derivation, and mention that transferring the solid must be done quickly to prevent heat loss to the air.

Question 43. Why is there a large temperature difference between land and sea that leads to land and sea breezes? Explain with reference to specific heat capacity.
Answer: Water has a very high specific heat capacity (\( 4200\text{ J kg}^{-1}\text{ K}^{-1} \)), which is approximately five times larger than that of dry sand or soil. Because of this, water takes a much longer time to heat up during the day and a correspondingly longer time to cool down at night compared to land. This significant difference in heating and cooling rates creates large temperature and pressure differences between the land and the sea, driving the convection currents known as land and sea breezes.
In simple words: Land heats up and cools down much faster than the sea because water has a high specific heat capacity. This difference in temperature causes breezes.

Exam Tip: Explicitly compare the specific heat capacity values of water and sand/soil to provide a robust physical explanation.

Question 44. State the principle of calorimetry.
Answer: The Principle of Calorimetry states that when a warmer body is mixed or placed in thermal contact with a colder body, heat energy is transferred from the warmer body to the colder body until they reach a common temperature, such that:
\[ \text{Total heat gained by the colder body} = \text{Total heat lost by the warmer body} \]
This holds true under the assumption that no heat energy is lost to the surrounding environment.
In simple words: When you mix hot and cold things, the heat lost by the hot thing is exactly equal to the heat gained by the cold thing, as long as no heat escapes.

Exam Tip: Always state the condition "assuming no heat loss to the surroundings" as the principle is only valid under this ideal state.

Question 45. Why is water used as an effective coolant in engines?
Answer: Water is widely used as a highly effective coolant because of its exceptionally high specific heat capacity (\( 4200\text{ J kg}^{-1}\text{ K}^{-1} \)). This property allows water to absorb massive amounts of heat from working engine components while undergoing only a relatively small increase in its own temperature.
In simple words: Water is a great coolant because it can absorb a lot of heat without getting too hot itself.

Exam Tip: This is a common application-based question. Make sure to specify the value of water's specific heat capacity to strengthen your argument.