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.

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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

Class 11 Physics Study Guides:

NCERT Solutions for Class 11 Physics	Physics Formula	Physics Study Notes
Difference between in Physics	Relation between articles in Physics	SI Units in Physics
Derivations in Physics	Physics Constants	Class 11 Physics Notes

CBSE CLASS XII Related Questions

1.
A battery of emf \( E \) and internal resistance \( r \) is connected to a rheostat. When a current of 2A is drawn from the battery, the potential difference across the rheostat is 5V. The potential difference becomes 4V when a current of 4A is drawn from the battery. Calculate the value of \( E \) and \( r \).
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2.
In an intrinsic semiconductor, carrier’s concentration is \( 5 \times 10^8 \ \text{m}^{-3} \). On doping with impurity atoms, the hole concentration becomes \( 8 \times 10^{12} \ \text{m}^{-3} \).

[(a)] Identify (i) the type of dopant and (ii) the extrinsic semiconductor so formed.

[(b)] Calculate the electron concentration in the extrinsic semiconductor.
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3.
An electric dipole consists of charges \(\pm 4 \mu C\) separated by a distance of \(6\,cm\). Calculate the electric field at a point on the axial line at a distance \(20\,cm\) from its center.
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4.
Two point charges \( 5 \, \mu C \) and \( -1 \, \mu C \) are placed at points \( (-3 \, \text{cm}, 0, 0) \) and \( (3 \, \text{cm}, 0, 0) \), respectively. An external electric field \( \vec{E} = \frac{A}{r^2} \hat{r} \) where \( A = 3 \times 10^5 \, \text{V m} \) is switched on in the region. Calculate the change in electrostatic energy of the system due to the electric field.
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5.
The ends of six wires, each of resistance R (= 10 \(\Omega\)) are joined as shown in the figure. The points A and B of the arrangement are connected in a circuit. Find the value of the effective resistance offered by it to the circuit.
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6.
The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is:
- \( \frac{\alpha}{r} \)
- \( \frac{\alpha r^2}{2} \)
- \( \frac{\alpha}{2r^2} \)
- \( -\frac{\alpha}{r} \)
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