SEE Science Light

1. A student sitting at the last bench of a class can read a book easily but wears spectacles to see the things written on the board. On the basis of this, answer the following questions.
1. What is the type of his defect?
2. What are the cause of defect of vision?
3. What type of spectacles is he using to remove his defect of vision.
4. Draw a diagram to show the remedy of this defect.
Ans: 1. He has short sighted eyes.
2. The inability of ciliary muscles to contract properly is its cause.
3. He is using spectacles with concave lens to remove his defect of vision.
4. The diagram to show the remedy of this defect is as follows:

2. What is short-sightedness?

Ans: The defect of vision in which a person can see clearly the near objects but cannot see distinctly the far objects is called short-sightedness.

3. What is long-sightedness?

Ans: The defect of vision in which a person can see the distant object clearly but cannot see distinctly the near objects is called long sightedness.

4. Study the given diagram and answer the following question;

I. What type of defect of vision of eyes is shown in the diagram?
II. What is the cause of such defect of vision?
III. What is done to remove such defect of vision?
IV. Draw the diagram to show the remedy of such defect of vision.
Ans: I. The type of defect of vision of eye shown in the diagram is short-sightedness or myopia. It is a defect of vision in which a person cannot see distant objects clearly, but can see nearby objects without any difficulty.

II. The causes behind this defect of vision are as followings:
a. The high conversing power of the eye lens or thickening of the eye lens.
b. If the eyeball gets elongated, the distance of the retina from the eye lens increases. It results in a decrease in the focal length of the eye lens.

III. To remove such defect of vision concave lens spectacles of suitable focal length are used.
IV.

Fig: Correction of Myopia using the concave lens.

5. A student sitting on the last bench of the class can see things written on the board but wears spectacles while reading a book. On the basis of this, answer the following questions:

I. What is the type of his defect?
II. What are the causes of a defect of vision?
III. What type of spectacles should be used to remove his defect of vision?
IV. Draw a diagram to show the remedy of this defect?
Ans:
I. The type of defect he has is hypermetropia or long sightedness. It is defined as the defect of vision in which a person cannot see nearby objects clearly but can see the distant object distinctly.
II. The causes of this type of defect of vision are as following:
a. As the eye lens becomes thin and the ciliary muscles fail to press the lens for the required thickness, it results in an increase in the focal length of the eye lens.
b. If the eyeball gets shortened, the distance between the eyes lens and retina becomes less. This also results in an increase in the focal length of the eye lens.
III. For the correction of this defect, a convex lens is suitable focal length is used in spectacles.
IV.

Fig: Use of the convex lens for the cure of hypermetropia.
6. What is accommodation?

Ans: The ability of the eye to focus on the object lying at various distances on the retina is called accommodation.

7. What is least distance of distinct vision?

Ans: The closest or least distance at which an object can be seen clearly is called least distance of distinct vision.

8. What is a telescope?

Ans: A telescope is an optical instrument used for seeing a distant object clearly and distinctly.

9. What is far point of the eyes?

Ans: The farthest point from where an object can be seen clearly is called a far point of the eyes.

10. How is the power of lens is calculated?

Ans: Power of lens is calculated by the formula P=1/f, where f is focal length of lens. It is negative for concave lens and positive for convex lens.

11. A convex lens has a focal length of 3 cm. If a candle is placed at twice of its focal length then:

I. Draw a clear ray diagram to show formation of image.
II. Write nature of image.
III. Calculate the magnification of image.
IV. What is the power of lens?
Ans: I. In the figure below, F is the principal focus of convex lens, O is the optical centre. Thus, the line OF represents focal length (f). The point 2F on the principal axis is at a distance of 2f from the optical centre O. If f = 3 cm, the point 2F is at a distance of 6 cm from O.

II. When an object is placed at twice of its focal length (2F), its image is formed at 2F on the other side of the object in the convex lens. The image shows following characteristics:
a) Image is real.
b) Image is inverted.
c) Image is same size as that of object.
III. In this case, the size of image is same as that of object. So, the magnification of the image is 1.
IV. Here,
Given,
Focal length (f) = 3 cm = 0.03m
Power (p) =?
We have,
P=1/f = 1/0.03 = 33.33 dioptres (D)
Hence, the power of lens is 33.33 D.
12. Study the diagram and answer the following questions.

I. Complete the following ray diagram.
II. Calculate the power of this lens.
III. Write any two nature of image formed by it.
Ans: I. In the figure below, F is the principal focus of the convex lens, O is the optical centre. Thus, the line OF represents focal length (f). The point 2F on the principal axis is at a distance of 2f from the optical centre O. If f = 2 cm, the point 2F is at a distance of 4 cm from O.
II. Do yourself.
III. The nature of image thus formed is,
real and inverted.

13. Complete the ray diagram given alongside. Also mention the nature of the image thus formed. If the power of hand lens is +3D, what should be the distance between the lens and book to read it properly?

Ans: In the figure below, F is the principal focus of the convex lens, O is the optical centre. Thus, the line OF represents focal length (f). The point 2F on the principal axis is at a distance of 2f from the optical centre O.
According to the question, the object is placed between focus (F) and optical Centre (O). When an object is placed between F and O, the convex lens forms an image on the same side as the object. The image is:
a) virtual
b) erect
c) magnified
Here,
Given,
Power (p) = +3D
The distance between hand lens and book to read the book properly
i.e. focal length (f) =?
We have,
f = 1/P = 1/3 = 0.3333 m = 33.33 cm
Hence, the distance between the hand lens and book should be 33.33 cm to read it properly.

14. What is a microscope?

Ans: A microscope is an optical instrument used to see tiny objects by forming their magnified images.

15. An object is placed at a distance of 20 cm from a convex lens of focal length 15 cm.a. Find the image distance
b. What is the power of the lens?
c. Find its magnification.
Ans: Do yourself

16. What is convex lens?

Ans: The lens which is thick at the middle and thin at the ends is called convex lens.

17. Why is convex lens is called a converging lens?

Ans: Convex lens is called a converging lens because it converges a parallel beam of light after refraction through it.

18. What is real image?

Ans: The image which can be obtained on the screen is called real image.

19. What is virtual image?

Ans: The image which cannot be obtained on the screen is called virtual image.

20. Draw a ray diagram to show the formation of image kept 2 cm far from the convex lens with focal length of 4 cm. Mention the nature of image with one application.

Ans: In the figure below, F is the principal focus of the convex lens, O is the optical centre. Thus, the line OF represents focal length (f). The point 2F on the principal axis is at a distance of 2f from the optical centre O. If f = 4 cm, the point 2F is at a distance of 10 cm from O.

According to the question, the object is placed between focus (F) and optical centre (O). When an object is placed between F and O, the convex lens forms an image on the same side as the object. The image is:
• virtual
• erect
• magnified
This type of position of the object in the convex lens is used in the simple microscope.

21. What is concave lens?

Ans: The lens which is thin at the middle and thick at the ends is called concave lens.
22. Draw ray for the formation of the image if an object is placed before a concave lens at the following positions and also write the nature of the image.

I. The object is placed at infinity.
II. The object is placed between F and 2F.
Ans: I. In the figure below, F is the principal focus of the concave lens, O is the optical centre. Thus, the line OF represents focal length (f). The point 2F on the principal axis is at a distance of 2f from the optical centre O.
When the object is at infinity, the image is formed at the focus (F). The characteristics of the image are as followings:
a. The image is virtual.
b. The image is erect.
c. The image is highly diminished.

Fig: The ray diagram for the concave lens when an object is placed at infinity.
Where,
O = Optical centre
F = Focus
II. In the figure below, F1 is the principal focus of the concave lens, O is the optical centre. Thus, the line OF represents focal length (f). The point 2F1 on the principal axis is at a distance of 2f from the optical centre.
When the object is placed between F1and 2F1 of a concave lens the image is formed between focus (F1) and the optical centre. The image is: The characteristics of the image are as followings:
a. The image is virtual.
b. The image is erect.
c. The image is diminished.

Fig: The ray diagram for a concave lens when an object is placed between F and 2F.
23. What is photographic camera?

Ans: Photographic camera is a simple optical instrument by means of which a permanent image of an object can be obtained on a photographic film.

24. What is light?

Ans: Light is a form of energy which produces the sensation of vision in our eyes.

25. What is an optical instrument?

Ans: An optical instrument is an arrangement, which makes use of a combination of lenses, mirrors, or prisms.

26. If focal length of lens of a spectacle is 50 cm, what will be its power? Assume that the spectacle is made up of concave lens.

Ans: Do yourself.
27. If power of a lens is -2.5 D, what is the focal length of the lens?

Ans: Do yourself.
28. An object is placed infront of convex lens at a distance twice its focal length. If the focal length of the lens is 4 cm, draw a neat ray diagram and write the nature of the image formed. Calculate the magnification. What is the power of this lens?

Ans: Do yourself.
29. How is the focal length of lens calculated using the lens formula?

Ans: Do yourself.

30. How is the magnification of lens is calculated?

Ans: Magnification is calculated by the formula m= I/O or v/u.

Where, I is size of image, O size of object, v image distance and u object distance.

31. What is lens?

Ans: Lens is a portion of a transparent refracting medium, usually glass, bound either by two curved surfaces or by one plane and one curved surface.
32. Why concave lens is called a diverging lens?

Ans: Concave lens is called a diverging lens because it diverges a parallel beam of light after refraction through it.

33. What is principal axis?

Ans: The straight line passing through the optical centre and perpendicular to the optical plane is known as principal axis.

34. What is principal focus?

Ans: The point on the principal axis where rays of light parallel to the principle axis after refraction through the lens, converge to a point on the axis or appears to diverge from a point on the axis is called principal focus.

35. What is focal length?

Ans: The distance between the optical center and the principal focus of a lens is known as the focal length.

36. What is focusing?

Ans: Changing the lens position so as to obtain the sharp image on the screen is called focusing.

37. What is power of lens?

Ans: Power of lens is defined as the capacity of the lens to converge or diverge the rays of light incident on it
38. What is range of vision?

Ans: The distance between the near point and the far point is called the range of vision.

39. Draw a ray diagram for the formation of image if an object is placed before a convex lens at the following position and mention the nature of the image.

I. At 2F
II. Between F and O
Ans: I. In the figure below, F₁ is the principal focus of a convex lens, O is the optical centre. Thus, the line OF₁ represents the focal length (f). The point 2F₁on the principal axis is at a distance of 2f from the optical centre O.
When an object is placed at 2F₁, its image is formed at 2F₂ on the other side in the convex lens. The image shows the following characteristics
a. An image is real.
b. An image is inverted.
c. An image is the same size as that of an object.

Fig: The ray diagram for a convex lens when an object is placed at 2F
Where, AB = Object
A'B' = Image
OB= Image distance
OB'= Object distance
O = Optical centre
OF₁= Focal length
II. In the figure below, F₁ is the principal focus of a convex lens, O is the optical centre. Thus, the line OF₁ represents the focal length (f). The point 2F₁on the principal axis is at a distance of 2f from the optical centre O.
When an object is placed between F₁ (Focus) and O (Optical centre), the convex lens forms an image on the same side as the object. The image shows the following characteristics
a. An image is virtual.
b. An image is erect.
c. An image is magnified.

Fig: The ray diagram for a convex lens when an object is placed in between F and O.
Where, AB = Object
A'B' = Image
OB = Image distance
OB' = Object distance
0 = Optical centre
OF₁ = Focal length
F₁ = Focus

40. Complete the given diagram and answer the following questions.

I. Complete the given ray diagram.
II. Which type of lens is used in the diagram?
III. Calculate the power of the lens.
IV. Write any two nature of image formed by it.
Ans:
I. In the figure below, F is the principal focus of the concave lens, O is the optical centre. Thus, the line OF represents focal length (f). The point 2F on the principal axis is at a distance of 2f from the optical centre O. If f = 2 cm, the point 2F is at a distance of 4 cm from O.
II. A concave lens is used in the diagram.
III. Do yourself.
41. What is magnification?

Ans: Magnification is defined as the ratio of the size of the image to the size of the object.

42. A convex lens has a focal length of 2 cm. An object kept at a distance of 6 cm from the lens forms an image 3 cm from it. Draw a ray diagram to show the formation of an image. What are the nature and magnification of the image? What is the power of the lens?

Ans:

Numerical part: Do yourself.

Nabraj Awasthi

SEE Science Heat

1. What is anomalous expansion of water?

Ans: The property of water in which water contracts on heating from 0^oC to 4^oC and then expands on further heating from 4^oC is called anomalous expansion of water.

2. Why does water overflow from the beaker when the beaker full of water at 4^oC is heated?
Ans: This is because water at 4^oC has minimum volume and when the beaker full of water at 4^oC is heated, water expands for which it overflows from the beaker.

3. If a 2kW electric immersion heater is used to warm 200 kg of water at 200C, calculate
a. The quantity of heat energy given out by immersion of heat in 2 hours’ time
b. Final temperature of water.

(Ans: 1440000J and 37.14°C)
4. On what factors does the quantity of heat of an object depend? How much heat is required to raise the temperature of 500 gm. of water from 5°C to 30°C?

Ans: We have the heat equation,

Q = msdt
where,
m = Mass of body
Q = Heat energy of a body
s = Specific heat capacity of body
dt = change in temperature
Hence, quantity of heat energy of an object depends in following factors:
• Mass of object
• Specific heat capacity of object
• Change in temperature
Here,
Given,
For water,
Mass (m) = 500 gm. = 0.5 kg
Initial temperature (t₁) = 5°C
Final temperature (t₂) = 30°C
Specific heat capacity (s) = 4200 J/kg°C
Heat energy (Q) =?
According to the heat equation,
Q =msdt
where,
m = Mass of body
Q = Heat gained or lost by a body
s = Specific heat capacity of the body
dt = Change in temperature
Q = 0.5 х 4200 х (30 — 5) = 52,500 J
Hence, the heat energy required to raise the temperature of 500 gm. water from 5°C to 25°C is 5.25х10⁴ J.
5. What do you mean by the statement that the specific heat capacity of water is 1 Cal/gm^oC?

Ans: The statement means that 1 gm of water requires 1 Calorie heat energy to raise its temperature by 1^oC.
6. What do you mean by the statement that the specific heat capacity of iron is iron is 470 J/kg^oC?

Ans: The statement means that 1 kg of iron needs 470 J heat energy to increase its temperature by 1^oC.
7. Why is water used in hot water bag?

Ans: Water is used in hot water bag because water has high specific heat capacity and water in hot water bag remains hot for a longer time than any other liquid.

8. Calculate the final temperature when 2400 J of heat is given to an iron of mass 2 kg at 200C. (s_iron = 460 Jkg^{-1 °}C^-1)

(Ans: 22.61°C)

9. What is temperature?

Ans: Temperature is defined as the degree of hotness or coldness of a body.
According to the molecular theory, temperature is the average kinetic energy of the molecules of a body.

10. What is absolute zero?

Ans: The coldest possible temperature at which the average kinetic energy of the molecules becomes zero is called absolute zero.

11. On which factor does the temperature of a body depend?

Ans: The temperature of a body depends on the average kinetic energy carried by the molecules of the body.

12. A metal ball of mass 200 gram at 83°C is plunged into 300 gram of water at 30°C. If the final temperature of the mixture is 33°C, calculate the specific heat capacity of metal.

Ans: Here,
Given,
Final temperature of mixture (t) = 33°C
For metal ball,
Mass (m₁) = 200 gm. = 0.2 kg
Initial temperature (t₁) = 83°C
Specific heat capacity (s₁) =?
For water,
Mass (m₂) = 300 gm. = 0.3 kg
Initial temperature (t₂) = 30°C
Specific heat capacity (s₂) = 4200 J/kg°C
According to heat equation,
Q = msdt
where,
m = Mass of body
Q = Heat gained or lost
s = Specific heat capacity of body
dt = Change in temperature
Again,
According to principle of calorimetry, the amount of heat gained is equal to the amount of heat lost.
So,

Heat lost by metal ball = Heat gained by water
m₁s₁dt₁ = m₂s₂dt₂
Or, 0.2 х s₁ х (83 — 33) = 0.3 х 4200 х (33 — 30)
s₁ = 378 J/kg°C
Hence, the specific heat capacity of metal 378 J/kg°C.
13. Three liquids A, B, C of equal mass are kept in the same type of container and placed in the sun for 30 minutes. The increase in temperature is given in the table. Study the table and answer the following questions:

I. Which liquid has the highest specific heat capacity? Why?
II. If equal mass of all the liquid at same temperature cooled, which one will cool faster? Why?
Ans: The specific heat capacity of a body is defined as the amount of heat required to raise the temperature of unit mass of that body by 1°C or 1K.
Different substances have different specific heat capacities. The bodies, which have more specific heat capacity, change their temperature slowly and those, which have less specific heat capacity, change their temperature faster.

I. The same amount of heat is applied to all three liquids for the equal time interval and liquid A has the lowest increase in temperature. As we know that, the substance which has more specific heat capacity, change their temperature slowly. So, the liquid A has more specific heat capacity.

II. If the equal mass of all three liquids at the same temperature is cooled, liquid C will cool faster, because the same amount of heat is applied to all three liquids for the equal time interval and liquid C has the highest increase in the temperature. That means liquid C has the lowest specific heat capacity and we know that substance which has less heat capacity, change their temperature faster.
14. If 200 gm. of water at 40°C is mixed with 2 kg of water at 30°C. Find the final temperature of the mixed water.

Solution:
Here,
Given,
For hot water,
Mass (m₁) = 200 gm. = 0.2 kg
Initial temperature (t₁) = 40°C
For cold water,
Mass (m₂) = 2 kg
Initial temperature (t₂) = 30°C
Specific heat capacity (s) = 4200 J/kg°C
Final temperature of mixture (t) =?
According to the heat equation,
Q =msdt
where,
m = Mass of body
Q = Heat gained or lost
s = Specific heat capacity of the body
dt = Change in temperature
Again,
According to the principle of calorimetry, the amount of heat gained is equal to the amount of heat loss.
So,
Heat lost by hot water = Heat gained by cold water
m₁sdt₁ = m₂sdt₂
Or, 0.2 х 4200 х (40 — t) = 2 х 4200 х (t — 30)
t = 30.9°C
Hence, the final temperature of the mixed water is 30.9°C.
15. Hot water of mass 10 kg at 90°C cooled for taking bath by mixing 20 kg of water at 20°C. What is the final temperature of water? Specific heat capacity of water is 4200 J/kg°C? Neglect the heat taken by bucket?
Solution: Here,Given,For hot water,

Mass (m₁) = 10 kg
Initial temperature (t₁) = 90°C
For cold water,
Mass (m₂) = 20 kg
Initial temperature (t₂) = 20°C
Specific heat capacity (s) = 4200 J/kg°C
Final temperature of mixture (t) =?
According to the heat equation,
Q = msdt
where,
m = Mass of body
Q = Heat gained or lost
s = Specific heat capacity of body
dt = Change in temperature
Again,
According to the principle of calorimetry, the amount of heat gained is equal to the amount of heat lost.
So,
Heat lost by hot water = Heat gained by cold water
m₁sdt₁ = m₂sdt₂
Or, 10 х 4200 х (90 — t) = 20 х 4200 х (t — 20)
t = 43.33°C
Hence, the final temperature of the water is 43.33°C.
.
17.
I. The temperature of an iron nail of mass 2 kg is 50°C and water of mass 200 gm. is 80°C, then which one substance has more heat energy? Calculate (The specific heat capacity of water and iron is 4200 J/kg°C and 1470J/kg°C respectively).

II. If the iron nail is put in the water in which direction heat energy will flow and why?

Solution:

I.
Here,
Given,
For iron nail,
Mass (m₁) = 2 kg
Temperature (t₁) = 50°C
Specific heat capacity (s1) = 470 J/kg°C
Heat energy (Q₁) =?
For water,
Mass (m₂) = 200 gm. = .2 kg
Temperature (t₂) = 80°C
Specific heat capacity (s) = 4200 J/kg°C
Heat energy (Q₂) =?
According to the heat equation,
Q =mst
where,
m = Mass of body
Q = Heat energy of a body
s = Specific heat capacity of body
t = Temperature
Again,
For iron nail,
Q₁ =m₁s₁t₁= 2 х 470 х 50 = 47,000 J
For water,
Q₂ =m₂s₂t₂ = .2 х 4200 х 80 = 67,200 J
Hence, Water has more heat energy.
II.
If an iron nail is put in the water, heat energy will flow from water to iron. Because the heat energy always flows from the body having a higher temperature to a body having a lower temperature.
18. What is thermometer?
Ans: The thermometer is an instrument used for measuring a temperature of a body.
19. Water is not used as a thermometric liquid. Why?

Ans: Water is not used as a thermometric liquid because water has not uniform expansion as it contracts on heating from 0^oC to 4^oC and then expands on further heating from 4^oC.
20. Write any three differences between clinical and laboratory thermometer.

21. What is thermal heat capacity?

Ans: The amount of heat required to raise the temperature of a whole body or substance through 1°C is called thermal heat capacity.

22. Derive Principle of caloriemetry

Derivation:
The principle of caloriemetry states,

“When two bodies of different temperatures are brought together, the body at higher temperature looses heat energy whereas the body at lower temperature gains heat energy. The process will continue until they reach at thermal equilibrium.”

The heat may get lost in external environment. But if the heat is not lost to external environment, then heat lost by body at higher temperature is equal to heat gained by body at lower temperature.

i.e Heat lost = Heat gained

This process continues until both bodies are at same temperature.

Let the mass of a body at higher temperature is m₁, its specific heat capacity is s₁ and the temperature of body is t₁.

Similarly, let the mass of a body at lower temperature is m₂, its specific heat capacity is s₂and the temperature of body is t₂.

If both bodies are brought together, then let the final temperature of the mixture be t.

If dt and dt’ are difference in temperature of bodies at higher and lower temperature respectively, then their values will be

dt = (t_{1 ­}­– t­)

and

dt' = (t - t₂ ­­)

So, the amount of heat energy lost by a body at higher temperature will be

Heat lost = m₁s₁dt

Similarly, the amount of heat energy gained by a body at lower temperature will be

Heat gained = m₂s₂dt'

Do you remember the principle of caloriemetry?

According to the principle of caloriemetry,

Heat lost = Heat gained

or, m₁s₁dt = m₂s₂dt'

or, m₁s₁ (t_{1 ­}­– t­) = m₂s₂ (t _­­– t­₂)

23. What is heat?

Ans: Heat is a form of energy that gives us the sensation of warmth.
According to the molecular theory, heat is the sum of total kinetic energy of the molecules of a body.

24. What is one Calorie?

Ans: One Calorie is defined as the amount of heat required to raise the temperature of one gram of pure water by 1°C.

25. What is the principle of caloriemetry?

Ans: The amount of heat gained is equal to the amount of heat lost, this relation is called principle of caloriemetry.

26. What is heat equation?

Ans: The product of mass, specific heat capacity and change in temperature of a body is equal to the heat gained or heat loss, this relation is called the heat equation.

27. On which factors does the heat lost or gained by a body depend?

Ans: The heat lost or gained by a body depends on mass of the body, change in temperature of the body and the specific heat capacity of the body.

28. Steam causes more burning than boiling water. Why?

Ans: Steam causes more burning than boiling water because steam contain more heat energy than the boiling water though both of them have same temperature.

29. On which factors does the heat content of a body depend?

Ans: The heat content of a body depends on the number of molecules contained in the body and the kinetic energy carried by the molecules.

30. Give reason:

I. Water is used to cool the engine of vehicles.
II. A new quilt is felt warmer than old one.
Ans:

I. Water has very high specific heat capacity (4200 J/kg°C) as compared to other liquids. The substances which have more specific heat capacity change their temperature slowly. Thus, water can absorb large amount of heat from the hot engines without appreciable rise in temperature of water. So, it prevents the engines of vehicle being excessively hot. Due to this reason water is used to cool the engine of vehicles.

II. A quilt feels warm because the air trapped in between the cotton or woolen batting acts as an insulator and does not allow the heat of the body to escape. However, when the quilt gets old, the cotton/woolen batting gets compressed and air spaces decreases. It does not remain as good an insulator to heat as it was earlier. Hence, a new quilt is felt warmer than old one.
31. In the given diagram, 100 gm. water is kept in the first beaker and 100 gm. mustard oil is kept in the second beaker. Study the diagram and answer the following question:

I. Which one has more temperature, if an equal amount of heat is supplied to both objects in equal time and why?
II. Which one cool fast, if both have a temperature of 100°C and why?
III. Which conclusion can be drawn from this activity?

Ans: I. If an equal amount of heat is applied to both objects, mustard oil will have more temperature. Because it has lower specific heat capacity in comparison to water and the substance which has less specific heat capacity, change their temperature faster. So, the temperature of the mustard oil will increase faster in comparison to water.

II. The bodies, which have more specific heat capacity, change their temperature slowly and those, which have less specific heat capacity, change their temperature faster. If both have temperature 100°C, mustard oil will cool faster. Because it has less specific heat capacity in comparison to water.

III. The conclusion drawn from this activity is, 'Different substances have different specific heat capacities. The bodies, which have more specific heat capacity, change their temperature slowly and those, which have less specific heat capacity, change their temperature faster'.
32. How much heat is supplied to raise the temperature of 100 gram of water from 5^°C to 90^°C? (Specific heat capacity of water = 4200 Jkg^{-1 °}C^-1)

(Ans: 35700J)

33. How much heat is required to raise the temperature of an aluminium kettle of mass 400 gram through 90⁰C? (Specific heat capacity of aluminium = 899 Jkg^{-1 0}C^-1)

(Ans: 32364J)

34. If 50 kJ of heat is transferred to 10 kg of water, what is the rise in its temperature? (Specific heat capacity of water is 4200 Jkg^{-1 °}C^-1)

(Ans: 1.19°C)

35. Calculate the specific heat capacity of the alloy of which a pressure cooker of mass 1.5 kg is made of if the quantity of heat necessary to raise its temperature by 60^°C is 81kJ.

(Ans: 900 J/kg°C)

36. If 200 ml of tea at 900C is mixed with 10 ml of milk at 150C, what will be the final temperature of the mixture? Assume that mass of 1 ml tea = mass of 1 ml milk = 1 gram and specific heat capacity of milk equals that of tea.

(Ans: 86.43°C)

37. Derive Q = m × s × dt

Derivation:

let mass of a body is m and the heat required to deviate the body temperature by dt is Q.

Experimentally, it has been found that;

• Heat energy (Q) lost or gained by a body is directly proportional to the mass (m) of the body.

i.e. Q ∝ m ................ ¹

• Heat energy (Q) lost or gained by a body is directly proportional to the change in temperature (dt) of the body.

i.e. Q ∝ dt ..................... ²

Combining relations 1 and 2, we get,

Q ∝ m × dt

The heat energy changes by certain multiplying factor, Specific heat capacity of the body.

The value of 'm' is different for different bodies.

Hence, the relation becomes

Q = m × s × dt

Change in body temperature (dt) is given by the difference between two temperature of a body.

If t­₁ and t₂ are initial and final temperatures of a body, then change in temperature is dt = t₂ ­– t­₁

• If the value of t₂ is greater than t₁, then the value of dt is positive.

i.e. the body temperature has increased. So, it must mean that the body absorbed heat energy.

• If the value of t₁ is greater than t₂, then the value of dt is negative.

i.e. the body temperature has decreased. So, it must mean that the body lost its heat energy.

The value of energy is always positive. So if a body loose heat energy, the change in temperature is dt = t₁– t­₂.

38. Specific heat capacities of three substances are given in the table. Answer the following questions observing the given table.

a) The specific heat capacity of substance ‘A’ is 910 J/kg°C, what does it mean?

b) Which one will gain the least temperature if equal heat is given to the equal masses of the three substances at equal temperature?

c) If equal masses of all three substances are heated upto 100°C and then kept on a wax slab, which one will go to the maximum depth in the wax slab?

d) If all of the substances are liquids, which one is suitable for using as a thermometric liquid?

Ans:

a) The specific heat capacity of the substance ‘A’ is 910 J/kg°C.

It means that we need to supply 910 Joule of heat to raise the temperature of 1 kg mass of substance ‘A’ at 1°C.

b) The substance having high specific heat capacity needs more heat to raise its temperature.

So, raise in temperature is slower in the substance having higher specific heat capacity.

Here, substance ‘C’ has the highest specific heat capacity.

Therefore, substance ‘C’ will gain the least temperature if equal heat is given to the equal masses of the three substances at equal temperature.

c) The substance having high specific heat capacity needs to lose more heat to fall its temperature.

So, cooling is slower in the substance having higher specific heat capacity.

Here, substance ‘C’ has the highest specific heat capacity.

Therefore, substance ‘C’ will remain hot for a long time when placed on the wax slab.

It can supply more heat to the wax during the cooling process and melts more wax.

So, substance ‘C’ will go to the maximum depth in the wax slab if equal masses of all three substances are heated up to 100°C and then kept on a wax slab.

d) The substance having less specific heat capacity gets heated faster with less heat.

So, it expands faster on heating.

The thermometric liquid should be highly heated sensitive.

It should expand quickly while measuring the temperature.

Here, substance “B” has the least specific heat capacity.

It expands faster than the other substances.

So, liquid “B” is suitable for use as a thermometric liquid.

39. What is specific heat capacity?

Ans: Specific heat capacity of a substance is defined as the amount of heat required to raise the temperature of 1 kilogram of the substance through 1°C.