physics ncert solutions class 11th Physics Class 11th Thermal Properties of Matter

11.19 Explain why :

(a) A body with large reflectivity is a poor emitter

(b) A brass tumbler feels much colder than a wooden tray on a chilly day

(c) An optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace

(d) The earth without its atmosphere would be inhospitably cold

(e) Heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water

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11.19 (a) A body with a large reflectivity is a bad absorber. A bad absorber will in turn be a poor emitter of radiations.

(b) Brass is a good conductor of heat, when one touches a brass tumbler, heat is conducted from the body to the brass tumbler easily. Hence, the temperature of the body reduces to a colder value and one feels cold.

On the other hand, wood is a poor conductor of heat. Very little heat is conducted from the body to the wooden tray. Resulting in negligible drop in body temperature.

Thus a brass tumbler feels colder than a wooden tray on a chilly day.

(c) Black body radiation equation is given by:

E = $σ$ ( $T^{4}$ &nbs

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11.19 (a) A body with a large reflectivity is a bad absorber. A bad absorber will in turn be a poor emitter of radiations.

On the other hand, wood is a poor conductor of heat. Very little heat is conducted from the body to the wooden tray. Resulting in negligible drop in body temperature.

Thus a brass tumbler feels colder than a wooden tray on a chilly day.

(c) Black body radiation equation is given by:

E = $σ$ ( $T^{4}$ - $T_{o}^{4}$ ), where

E = energy radiation

$σ$ = constant

$T =$ temperature of the optical pyrometer

$T_{o}$ = Temperature of open space

$H e n c e a n i n c r e a s e i n t h e t e m p e r a t u r e o f o p e n s p a c e r e d u c e s t h e r a d i a t i o n e n e r g y$

When the same piece of iron placed in a furnace, the radiation energy is E = $σ T^{4}$

(d) In the absence of atmospheric gases, no extra heat will be trapped. All the heat would be radiated back from Earth’s surface.

(e) A heating system based on the circulation of steam is more efficient in warming a building than that based on the circulation of hot water. This is because steam contains surplus heat in the form of latent heat ( 540 cal/g)

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11.19 (a) A body with a large reflectivity is a bad absorber. A bad absorber will in turn be a poor emitter of radiations.(b) Brass is a good conductor of heat, when one touches a brass tumbler, heat is conducted from the body to the brass tumbler easily. Hence, the temperature of the body reduces to a colder value and one feels cold.On the other hand, wood is a poor conductor of heat. Very little heat is conducted from the body to the wooden tray. Resulting in negligible drop in body temperature.Thus a brass tumbler feels colder than a wooden tray on a chilly day.(c) Black body radiation equation is given by:E = <math><mi>σ</mi></math> ( <math><msup><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>4</mn></mrow></mrow></msup></math> - <math><msubsup><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mi>o</mi></mrow></mrow><mrow><mrow><mn>4</mn></mrow></mrow></msubsup></math> ), whereE = energy radiation<math><mi>σ</mi></math> = constant<math><mi>T</mi><mo>=</mo><mi></mi></math> temperature of the optical pyrometer<math><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mi>o</mi></mrow></mrow></msub></math> = Temperature of open space <math> <mi mathvariant="normal">H</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal">c</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">i</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal">c</mi> <mi mathvariant="normal">r</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">s</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">i</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">t</mi> <mi mathvariant="normal">h</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">t</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">m</mi> <mi mathvariant="normal">p</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">r</mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">t</mi> <mi mathvariant="normal">u</mi> <mi mathvariant="normal">r</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">f</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">p</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">s</mi> <mi mathvariant="normal">p</mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">c</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">r</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">d</mi> <mi mathvariant="normal">u</mi> <mi mathvariant="normal">c</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">s</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">t</mi> <mi mathvariant="normal">h</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">r</mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">d</mi> <mi mathvariant="normal">i</mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">t</mi> <mi mathvariant="normal">i</mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">r</mi> <mi mathvariant="normal">g</mi> <mi mathvariant="normal">y</mi> </math> When the same piece of iron placed in a furnace, the radiation energy is E = <math> <mi>σ</mi> <msup> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </msup> </math> (d) In the absence of atmospheric gases, no extra heat will be trapped. All the heat would be radiated back from Earth’s surface.(e) A heating system based on the circulation of steam is more efficient in warming a building than that based on the circulation of hot water. This is because steam contains surplus heat in the form of latent heat ( 540 cal/g)

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At what temperature a gold ring of diameter 6.230cm be heated so that it can be fitted on a wooden bangle of diameter. 6.241cm? Both the diameters have been measured at room temperature (27°C).

(Given : coefficient of linear thermal expansion of gold a_L = 1.4 × 10^-⁵ K^-¹)

alok kumar singh

According to question, we can write

$d = d_{0} (1 + α Δ T) \Rightarrow 6.241 = 6.230 (1 + 1.4 \times 1 0^{- 5} \times Δ T)$

$\Rightarrow T = 126.18 + 27 = 153.18 ° C$

A copper block of mass 5.0kg is heated to a temperature of $500 ° C$ and is placed on a large ice block. What is the maximum amount of ice that can melt?

[Specific heat of copper : 0.39 J g^-1 ⁰C^-1 and latent heat of fusion of water : 335 J g^-1]

Raj Pandey

Heat Released by block = Heat gain by large Ice block

$\Rightarrow m c Δ T = M_{i c e} L$

5 × 0.39 × 500 = m_ice × 335

$m_{i c e} = \frac{5 \times 0.39 \times 500}{335}$

= 2.91 kg

A solid metallic cube having total surface area 24 m² is uniformly heated. if its temperature is increased by 10⁰C, calculate the increase in volume of the cube.

(Given a = 5.0 × 10^-4 ⁰C^-1).

Raj Pandey

Given

6a² = 24

a2 = 4

a = 2m

$Δ v = v (γ) Δ T$

$= 3 α v Δ T$

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A glass full of hot milk is poured on the table. It begins to cool gradually. Which of the following is correct?

(a) The rate of cooling is constant till milk attains the temperature of the surrounding

(b) The temperature of milk falls off exponentially with time

(c) While cooling, there is a flow of heat from milk to the surrounding as well as from surrounding to the milk but the net flow of heat is from milk to the surrounding and that is why it cools

(d) All three phenomenon, conduction, convection and radiation are responsible for the loss of heat from milk to the surroundings

Payal Gupta

This is a multiple choice answer as classified in NCERT Exemplar

(b), (c), (d) When the hot milk in the table is transferred to the surroundings by conduction, convection and radiation.

According to newton's law of cooling temperature of the milk falls of exponentially. Heat also will be transferred from surroundings to the milk but will be lesser than that of transferred from milk to surroundings. So option b, c, d satisfy.

Refer to the plot of temperature versus time (Fig) showing the changes in the state of ice on heating (not to scale) Which of the following is correct?

(a) The region AB represents ice and water in thermal equilibrium

(b) At B water starts boiling

(c) At C all the water gets converted into steam

(d) C to D represents water and steam in equilibrium at boiling point

Payal Gupta

This is a multiple choice answer as classified in NCERT Exemplar

(a), (d) During the process AB temperature of the system is 0⁰C . hence it represents phase change that is transformation of ice into water while temperature remains 0⁰C.

BC represents rise in temperature of water from 0⁰C to 100⁰C. now water starts converting into steam which is represented by CD.

Which is only represents by a, d options

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