| Publisher | : Santra publication pvt. ltd. |
| Book Name | : Madhyamik Physical Science And Environment |
| Class | : 9 (Madhyamik) |
| Subject | : Physical Science |
| Chapter Name | : Heat |
Multiple Choice Questions (MCQ)
[Each of Mark-1]
Question 1
The volume of water in the temperature range from 0°C to 4°C
- increases
- remains the same
- decreases
- may either increase or decrease
Answer
decreases
Explanation
The volume of water decreases within the temperature range from 0°C to 4°C due to the phenomenon of thermal contraction.
Question 2
The coefficient of expansion of water in the range 0°C to 4°C is
- zero
- negative
- positive
- infinity
Answer
negative
Explanation
The coefficient of expansion of water in the range of 0°C to 4°C is negative, indicating that water contracts as its temperature decreases.
Question 3
Due to the special behaviour of water, lakes, and ponds freeze
- first at the bottom
- first in the middle layer
- first at their surface
- anywhere
Answer
first at their surface
Explanation
Lakes and ponds freeze first at their surface due to the lower density of ice compared to liquid water, allowing it to float.
Question 4
The amount of heat required to raise the temperature of a body depends on
- the mass only
- the temperature
- the specific heat only
- the all three of them
Answer
the all three of them
Explanation
The amount of heat required to raise the temperature of a body depends on all three factors: the mass, the temperature change, and the specific heat of the substance.
Question 5
By definition 1 cal is equal to
- 1 J
- 14.186 J
- 4.186 J
- none
Answer
4.186 J
Explanation
By definition, 1 cal (calorie) is equal to 4.186 J (joules), which is the amount of energy required to raise the temperature of 1 gram of water by 1 degree Celsius.
Question 6
For m = mass of calorimeter and s = specific heat or the calorimeter material, the water – equivalent of the calorimeter is
- m
- n
- m2s
- ms
Answer
ms
Explanation
The water equivalent of the calorimeter is given by the product of the mass of the calorimeter (m) and its specific heat (s), i.e., ms.
Question 7
Water boils at a temperature (when pressure is 1.5 atm)
- T = 100°C
- T > 100°C
- T < 100°C
- none
Answer
T > 100°C
Explanation
Water boils at a temperature higher than 100°C when the pressure is 1.5 atm. The boiling point of water increases with increasing pressure.
Question 8
The heat of vaporization of water at 100°C is
- 4.186 J
- 539 J/g
- 539 cal/g
- 80 cal/g
Answer
539 cal/g
Explanation
The heat of vaporization of water at 100°C is approximately 539 cal/g or 2260 J/g. This represents the energy required to convert water from liquid to vapour at that temperature.
One/two-word or single-sentence answer questions
[Each of Mark-1]
Question 1
Mention the unit (SI) of heat.
Answer
Joule (J) is the SI unit of heat.
Question 2
Which temperature is higher 30°C or 300 K?
Answer
300 K is higher than 30°C since the Kelvin scale starts at absolute zero and has the same unit size as the Celsius scale.
Question 3
What is the value of specific heat of water?
Answer
The specific heat of water is approximately 4.186 J/g°C.
Question 4
What is the SI unit of specific heat?
Answer
The SI unit of specific heat is joule per gram per degree Celsius (J/g°C).
Question 5
What do you mean by 1 calorie of heat?
Answer
One calorie of heat refers to the amount of heat energy required to raise the temperature of 1 gram of water by 1 degree Celsius.
Question 6
What is heat capacity?
Answer
Heat capacity refers to the amount of heat energy required to raise the temperature of an object or a substance by 1 degree Celsius.
Question 7
What is specific heat?
Answer
Specific heat refers to the amount of heat energy required to raise the temperature of a unit mass of a substance by 1 degree Celsius (or 1 Kelvin).
Question 8
Lead has a specific heat of 0.03. What does it mean?
Answer
A specific heat of 0.03 for lead means that it requires 0.03 joules of heat energy to raise the temperature of 1 gram of lead by 1 degree Celsius (or 1 Kelvin).
Question 9
What is the value of heat of fusion of ice?
Answer
The heat of fusion of ice is approximately 336 J/g or 80 cal/g.
Question 10
What is the value of heat of vaporization of water?
Answer
The heat of vaporization of water is approximately 40.7 kJ/mol or 2260 J/g
Short answer questions
[Each of Mark-2]
Question 1
Do same amount of iron and copper at the same temperature hold same amount of heat? Explain
Answer
No, the same amount of iron and copper at the same temperature do not hold the same amount of heat. The heat capacity of a substance depends on its specific heat and mass. Since iron and copper have different specific heat values, they will require different amounts of heat energy to achieve the same temperature change.
Question 2
State and explain the basic principle of calorimetry.
Answer
The basic principle of calorimetry is that the heat lost by a hot body is equal to the heat gained by a cold body when they are placed in contact in an insulated system. This is a consequence of the law of conservation of energy.
In short, heat lost = heat gained.
This principle is used to measure the heat capacity of materials and the enthalpy of reactions.
Question 3
Mention at least three points to distinguish between heat and temperature.
Answer
| Heat | Temperature |
|---|---|
| Quantity of energy | Measure of hotness |
| Depends on mass | Independent of mass |
| Transferred between objects | Property of an object |
Question 4
Why water is used in a hot (water) bag? Explain.
Answer
Water is used in a hot water bag because it has a high specific heat capacity, which allows it to absorb and retain heat efficiently. This makes it suitable for storing and releasing heat gradually, providing sustained warmth when used in a hot water bag.
Question 5
Write two differences between thermal capacity and water equivalent.
Answer
| Thermal Capacity | Water Equivalent |
|---|---|
| Depends on specific heat and mass | Depends only on the mass |
| Represents heat energy required for temperature change | Represents mass of water with equivalent heat absorption capacity |
| Applicable to any substance | Applicable only to objects with significant water content |
Question 6
Write the SI units of thermal capacity and water equivalent.
Answer
The SI unit of thermal capacity is joule per degree Celsius (J/°C), and the SI unit of water equivalent is kilogram (kg).
Question 7
What is the significance of the mechanical equivalent of heat 4.2 × 107 erg/cal?
Answer
The mechanical equivalent of heat, 4.2 × 107 erg/cal, is significant as it establishes the relationship between mechanical energy and heat energy, allowing for the conversion and understanding of energy in different forms.
Question 8
What are the factors that the amount of heat absorbed or released by a body depends on?
Answer
The amount of heat absorbed or released by a body depends on the following factors:
- Mass: The greater the mass of the body, the more heat energy it can absorb or release.
- Temperature Change: The larger the change in temperature experienced by the body, the greater the amount of heat absorbed or released.
- Specific Heat Capacity: Each substance has a specific heat capacity, which represents the amount of heat energy required to raise the temperature of a given mass of the substance by a specific amount.
Question 9
Define and explain the thermal capacity of a body.
Answer
Thermal capacity refers to the ability of a body to store heat energy. It is the amount of heat energy required to raise the temperature of the body by 1 degree Celsius. It depends on the mass and specific heat capacity of the body. A body with higher thermal capacity can store more heat energy and will experience a smaller temperature change for a given amount of heat input or output.
Question 10
Find out the water equivalent and thermal capacity of a calorimeter made of 100 g of copper (s = 0.09 cal/g/°C). [Ans. 9 g, 9 cal/°C]
Answer
To find the water equivalent and thermal capacity of the calorimeter, we need to use the formula:
Water Equivalent = Mass of the calorimeter × Specific heat of the calorimeter
Thermal Capacity = Water Equivalent × Specific heat of water
Given:
Mass of the calorimeter (m) = 100 g
Specific heat of the calorimeter (s) = 0.09 cal/g/°C
Water Equivalent = 100 × 0.09 = 9 g
Thermal Capacity = 9 × 1 = 9 cal/°C
Therefore, the water equivalent of the calorimeter is 9 g, and its thermal capacity is 9 cal/°C.
Long answer question
[Each of Mark-2/3]
Question 1
Distinguish between thermal capacity and water equivalent.
Answer
| Thermal Capacity | Water Equivalent |
|---|---|
| Depends on specific heat and mass | Depends only on the mass |
| Represents heat energy required for temperature change | Represents mass of water with equivalent heat absorption capacity |
| Applicable to any substance | Applicable only to objects with significant water content |
Question 2
50.4 joule of work is needed to produce 12 cal of heat. Find the value of Joule’s constant. [Ans. 4.2 J/cal]
Answer
To find the value of Joule’s constant, we can use the equation:
Joule’s constant (J) = Work done (in joules) / Heat produced (in calories)
Given:
Work done = 50.4 joules
Heat produced = 12 cal
J = 50.4 J \over 12 cal
Converting calories to joules: 1 cal = 4.186 J
J = 50.4 \over(12 × 4.186)
J ≈ 4.2 J/cal
Therefore, the value of Joule’s constant is approximately 4.2 J/cal.
Question 3
How much heat to be supplied to raise the temperature of 100 g of water from 20°C to the normal boiling point of water? [Ans. 8000 cal]
Answer
To calculate the heat required to raise the temperature of 100 g of water from 20°C to the normal boiling point (100°C), we can use the formula:
Heat = Mass × Specific heat × Temperature change
Given:
Mass of water (m) = 100 g
Specific heat of water (s) = 1 cal/g/°C
Temperature change (ΔT) = (100°C – 20°C) = 80°C
Heat = 100 × 1 × 80
Heat = 8000 cal
Therefore, the amount of heat that needs to be supplied to raise the temperature of 100 g of water from 20°C to the normal boiling point is 8000 cal.
Question 4
What temperature reads the same in both the °F and °C scales? [Ans. -40]
Answer
The temperature -40 degrees Fahrenheit (-40 °F) is the same as -40 degrees Celsius (-40 °C).
To explain why this is the case, we can use the conversion formula between Fahrenheit and Celsius:
°C = (°F – 32) × 5/9
Let’s substitute -40 °F into the formula:
°C = (-40 - 32) × 5\over 9
= –72 × 5\over 9 ≈ -40
So, -40 °C is the temperature that corresponds to -40 °F, making it the temperature that reads the same on both the Fahrenheit and Celsius scales.
Question 5
How much heat would be released by a 60 g iron body (specific heat = 0.12 cal/g/C) when its temperature is reduced from 200°C to 100°C? [Ans. 720 cal]
Answer
To calculate the heat released by the iron body when its temperature is reduced from 200°C to 100°C, we can use the formula:
Heat = Mass × Specific heat × Temperature change
Given:
Mass of iron body (m) = 60 g
Specific heat of iron (s) = 0.12 cal/g/°C
Temperature change (ΔT) = (200°C – 100°C) = 100°C
Heat = 60 × 0.12 × 100°C
Heat = 720 cal
Therefore, the amount of heat released by the 60 g iron body when its temperature is reduced from 200°C to 100°C is 720 cal.
Question 6
Consider a 3.0 kg copper body. How much heat is required to raise its temperature by 10°C? [s= 0.09 cal/g/°C]. [Ans. 2700 cal]
Answer
To calculate the heat required to raise the temperature of a 3.0 kg copper body by 10°C, we can use the formula:
Heat = Mass × Specific heat × Temperature change
Given:
Mass of copper body (m) = 3.0 kg
Specific heat of copper (s) = 0.09 cal/g/°C
Temperature change (ΔT) = 10°C
First, we need to convert the mass from kilograms to grams:
Mass = 3.0 × 1000 = 3000 g
Heat = 3000 × 0.09 × 10
Heat = 2700 cal
Therefore, the amount of heat required to raise the temperature of the 3.0 kg copper body by 10°C is 2700 cal.
Question 7
100 g of a copper body at 60°C is cooled to 40°C. How much heat does it release? [Ans. 780 cal]
Answer
To calculate the heat released by a 100 g copper body when it is cooled from 60°C to 40°C, we can use the formula:
Heat = Mass × Specific heat × Temperature change
Given:
Mass of copper body (m) = 100 g
Specific heat of copper (s) = 0.39 cal/g/°C
Temperature change (ΔT) = (40°C – 60°C) = -20°C (negative since it is a decrease in temperature)
Heat = 100 × 0.39 × -20
Heat = – 780 cal
The negative sign indicates that heat is being released by the copper body.
Therefore, the copper body releases approximately 780 cal of heat when it is cooled from 60°C to 40°C.
Question 8
Express -40°C in the °F scale. [Ans. -40°F]
Answer
To convert -40°C to the °F scale, we can use the conversion formula:
°F = (°C × 9/5) + 32
Let’s substitute -40°C into the formula:
°F = (-40 × 9/5) + 32 °F
= -72 + 32 °F = -40
Therefore, -40°C is equal to -40°F on the °F scale.
Question 9
Write the relation of the quality of heat with different factors that it depends on. Explain the terms.
Answer
The quality of heat refers to the usefulness or the ability of heat energy to perform work or produce a desired effect. It depends on several factors:
- Temperature: The higher the temperature of the heat source, the higher its quality. Heat at a high temperature has greater potential to do work or produce a desired effect compared to heat at a lower temperature.
- Heat Transfer Mechanism: The efficiency of the heat transfer mechanism affects the quality of heat. For example, if heat is transferred through a highly efficient process such as direct conduction or radiation, the quality remains high. In contrast, if heat is transferred through an inefficient process like convection or by low-grade heat sources, the quality may be lower.
- Specific Application: The suitability of heat for a specific application determines its quality. Heat energy needs to match the requirements of the system or process in which it is utilized. For instance, heat that is well-matched to the needs of an engine to produce mechanical work would be considered higher quality for that specific application.
Question 10
Define heat capacity and specific heat. Find their SI units.
Answer
Heat capacity refers to the amount of heat energy required to raise the temperature of an object by 1 degree Celsius. Its SI unit is joule per degree Celsius (J/°C).
Specific heat, on the other hand, refers to the amount of heat energy required to raise the temperature of a unit mass of a substance by 1 degree Celsius. Its SI unit is joule per gram per degree Celsius (J/g/°C).
