NCERT Solutions For Class 11 Physics Chapter 11: Thermal Properties of Matter

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NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter are given in the article. The matter is the one due to which a matter exhibits heat conductivity or it is the property that decides the nature of the matter in the presence of heat. Thus, thermal properties are exhibited by objects when heat passes through them. Temperature is one of the physical properties of matter.

Class 11 Physics Chapter 11 Thermal Properties of Matter belongs to Unit 7 Properties of Bulk Matter which along with Unit 8 and Unit 9 has a weightage of 20 marks. NCERT Solutions for Chapter 11 Class 11 Physics covers concepts of specific heat capacity, Latent heat formula, and Blackbody Radiation.

Download PDF: NCERT Solutions for Class 11 Physics Chapter 11

NCERT Solutions for Class 11 Physics Chapter 11

Class 11 Physics Chapter 11 – Concepts Covered

The thermal energy of a body is the quantity of heat that is required to raise the temperature of the whole body by a unit degree. If Q is the amount of heat that’s needed to produce a temperature change (Δt), then the thermal capacity of the substance is given as:

$S = {Q \over \bigtriangleup t}$

The Specific Heat Capacity of a substance is the amount of heat that is required to raise the temperature of a unit mass of a substance by 1° C.

$S = {1 \over m} {Q \over \bigtriangleup t}$

Calorimetry is concerned with the measurement of heat.

From the law of energy conservation: Heat gained by one body = heat lost by the other

The Basic Heat Formula: The heat Q that is required to raise the temperature of a substance with mass m of specific heat capacity s through t degrees is given by

Q = m x S x t

Newton’s law of cooling states that the rate of loss of heat in a body is proportional to the difference in temperature of the body and its surroundings provided that the difference in temperature is small and is not more than 40° C.

${dT \over dt} = -K(T-T_s)$

The negative sign implies that as time passes, the temperature decreases.

Class 11 Physics Chapter 11 Related Guides:

Specific Heat Capacity	Change of state	Latent Heat
Heat Transfer	Wien’s Displacement Law	Heat Capacity Formula
Thermal Expansion	Thermal Conductivity of Copper	Regelation

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NCERT Solutions for Class 11 Physics	Physics Formula	Physics Study Notes
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CBSE CLASS XII Related Questions

1.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:
- 5 V
- 3.5 V
- 2.5 V
- Zero V
View Solution
2.
Two point charges -Q and Q are located at points (d, 0) and (0, d) respectively, in x-y plane. The electric field E at the origin will be:
- $\frac{1}{4\pi\epsilon_0} \frac{\sqrt{2}Q}{d^2} (\hat{i} - \hat{j})$
- $\frac{1}{4\pi\epsilon_0} \frac{\sqrt{2}Q}{d^2} (-\hat{i} - \hat{j})$
- $\frac{1}{4\pi\epsilon_0} \frac{Q}{d^2} (\hat{i} - \hat{j})$
- $\frac{1}{4\pi\epsilon_0} \frac{Q}{d^2} (-\hat{i} - \hat{j})$
View Solution
3.
In a region, the electric potential varies as $V = 10 - 50x$, where $V$ is in volts and $x$ is in meters. The electric field in the region is
- $10\ \text{N/C}$ along $+x$
- $10\ \text{N/C}$ along $-x$
- $50\ \text{N/C}$ along $+x$
- $50\ \text{N/C}$ along $-x$
View Solution
4.
Draw the plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions that can be drawn from this plot.
View Solution
5.
A ray of light MN is incident normally on the face corresponding with side AB of a prism with an isosceles right-angled triangular base ABC. Trace the path of the ray as it passes through the prism when the refractive index of the prism material is (i) $\sqrt{2}$, and (ii) $\sqrt{3}$.
View Solution
6.
A parallel beam of monochromatic light falls normally on a single slit of width $a$ and a diffraction pattern is observed on a screen placed at distance $D$ from the slit. Explain:
(I) the formation of maxima and minima in the diffraction pattern, and
(II) why the maxima go on becoming weaker and weaker with its increasing number $n$.
View Solution

View All

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NCERT Solutions For Class 11 Physics Chapter 11: Thermal Properties of Matter

NCERT Solutions for Class 11 Physics Chapter 11

Class 11 Physics Chapter 11 – Concepts Covered

CBSE CLASS XII Related Questions

1.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:

2.
Two point charges -Q and Q are located at points (d, 0) and (0, d) respectively, in x-y plane. The electric field E at the origin will be:

3.
In a region, the electric potential varies as \(V = 10 - 50x\), where \(V\) is in volts and \(x\) is in meters. The electric field in the region is

4.
Draw the plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions that can be drawn from this plot.

5.
A ray of light MN is incident normally on the face corresponding with side AB of a prism with an isosceles right-angled triangular base ABC. Trace the path of the ray as it passes through the prism when the refractive index of the prism material is (i) $\sqrt{2}$, and (ii) $\sqrt{3}$.

Similar Physics Concepts

Comments

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NCERT Solutions For Class 11 Physics Chapter 11: Thermal Properties of Matter

NCERT Solutions for Class 11 Physics Chapter 11

Class 11 Physics Chapter 11 – Concepts Covered

CBSE CLASS XII Related Questions

1.A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:

2.Two point charges -Q and Q are located at points (d, 0) and (0, d) respectively, in x-y plane. The electric field E at the origin will be:

3.In a region, the electric potential varies as \(V = 10 - 50x\), where \(V\) is in volts and \(x\) is in meters. The electric field in the region is

4.Draw the plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions that can be drawn from this plot.

5.A ray of light MN is incident normally on the face corresponding with side AB of a prism with an isosceles right-angled triangular base ABC. Trace the path of the ray as it passes through the prism when the refractive index of the prism material is (i) $\sqrt{2}$, and (ii) $\sqrt{3}$.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:

2.
Two point charges -Q and Q are located at points (d, 0) and (0, d) respectively, in x-y plane. The electric field E at the origin will be:

3.
In a region, the electric potential varies as \(V = 10 - 50x\), where \(V\) is in volts and \(x\) is in meters. The electric field in the region is

4.
Draw the plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions that can be drawn from this plot.

5.
A ray of light MN is incident normally on the face corresponding with side AB of a prism with an isosceles right-angled triangular base ABC. Trace the path of the ray as it passes through the prism when the refractive index of the prism material is (i) $\sqrt{2}$, and (ii) $\sqrt{3}$.