Javascript required
Skip to content Skip to sidebar Skip to footer
Home / Ncert Solutions for Class 11 Physics Thermal Properties of Matter

Ncert Solutions for Class 11 Physics Thermal Properties of Matter

NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter: Temperature is a relative measure of indication of coldness (or hotness) while heat is the form of energy transferred between two or more systems or a system and its surroundings due to the temperature difference. The solutions of NCERT class 11 physics chapter 11 thermal properties of matter start with the questions related to temperature conversions from one scale to another. All the questions discussed in the CBSE NCERT solutions for class 11 physics chapter 11 thermal properties of matter are important to understand the concept studied in the chapter. NCERT solutions are an important tool to perform well in exams. Consider a day in the winter season. Let the temperature at 6 am is 5 degree and 9 am be 10 degrees. We will say that 6 am is much colder than 9 am or we can say 9 am is hotter than 6 am which means cold and hot are relative terms. We know that a glass of ice left on a table on a hot day eventually warms up whereas a cup of hot coffee on the same table cools down. It means that when the temperature of the body and its surrounding medium are different, heat transfer takes place between the system and the surrounding medium until the body and the surrounding medium are at the same temperature. So hot coffee becomes cold and cold water becomes warmer.

Questions based on the following topics are discussed in the NCERT solutions for class 11 physics chapter 11 thermal properties of matter.

11.2 Temperature and heat

11.3 Measurement of temperature

11.4 Ideal-gas equation and absolute temperature

11.5 Thermal expansion

11.6 Specific heat capacity

11.7 Calorimetry

11.8 Change of state

11.9 Heat transfer

11.10 Newton's law of cooling

The difference between temperature and heat can be well understood from the following graph given in NCERT .

Even though heat is increased during the phase change, the temperature remains constant until the phase change has occurred. That is increasing heat always does not imply that there is an increase in temperature.

NCERT solutions for class 11 physics chapter 11 thermal properties of matter exercise

Q. 11.1 The triple points of neon and carbon dioxide are 24.57\; K and 216.55\; K respectively. Express these temperatures on the Celsius and Fahrenheit scales.

Answer:

The relation between Kelvin and Celcius scale is T K = T C + 273.15

Triple Point of Neon in Kelvin T K = 24.57 K

Triple Point of Neon in Celcius T C = T K -273.15 = 24.57 -273.15 =-248.58 o C

Triple Point of carbon dioxide in Kelvin T K = 216.55 K

Triple Point of carbon dioxide in Celcius T C = T K -273.15 = 216.55 - 273.15 = -56.60 o C

The relation between Celcius and Fahrenheit scale is T_{F}=\frac{9}{5}T_{C} + 32

Triple Point of Neon in Fahrenheit is

T_{F}=\frac{9}{5}\times (-248.58) + 32

T F = -415.44 o C

Triple Point of carbon dioxide in Fahrenheit is

T_{F}=\frac{9}{5}\times (-56.60) + 32

T F = -69.88 o C

Thermal Properties of Matter Excercise:

Question:

Q. 11.15 Given below are observations on molar specific heats at room temperature of some common gases.

Gas Molar specific heat (Cv )

(cal\; mol^{-1}K^{-1})

Hydrogen 4.87

Nitrogen 4.97

Oxygen 5.02

Nitric oxide 4.99

Carbon monoxide 5.01

Chlorine 6.17

The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92\; cal/mol\; K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?

Answer:

Monoatomic gases have only translational degree of freedom but diatomic gases have rotational degrees of freedom as well. The temperature increases with increase in the spontaneity of motion in all degrees. Therefore to increase the temperature of diatomic gases more energy is required than that required to increase the temperature of monoatomic gases by the same value owing to higher degrees of freedom in diatomic gases.

If we only consider rotational modes of freedom the molar specific heat of the diatomic gases would be given as

\\c=\frac{fR}{2}\\ c=\frac{5}{2}\times 1.92\\ c=4.95\ cal\ mol^{-1}\ K^{-1}

The number of degrees of freedom = 5 (3 translational and 2 rotational)

The values given in the table are more or less in accordance with the above calculated one. The larger deviation from the calculated value in the case of chlorine is because of the presence of vibrational motion as well.

Q. 11.19 (d) Explain why :

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

Answer:

The sun rays contain infrared radiations. These are reflected back by the lower part of the atmosphere after being reflected by the surface of the earth and are trapped inside the atmosphere thus maintaining the Earth's temperature at a hospitable level. Without these rays being trapped the temperature of the earth will go down severely and thus the Earth without its atmosphere would be inhospitably cold.

Q. 11.20 A body cools from 80^{\circ}C to 50^{\circ}C in 5 minutes. Calculate the time it takes to cool from 60^{\circ}C to 30^{\circ}C. The temperature of the surroundings is 20^{\circ}C.

Answer:

Let a body initially be at temperature T 1

Let its final Temperature be T 2

Let the surrounding temperature be T 0

Let the temperature change in time t.

According to Newton's Law of cooling

\\-\frac{dT}{dt}=K(T-T_{0})\\ \frac{dT}{T-T_{0}}=-Kdt\\ \int_{T_{1}}^{T_{2}}\frac{dT}{T-T_{0}}=-K\int_{0}^{t}dt\\ \left [ ln(T-T_{0}) \right ]_{T_{1}}^{T_{2}}=-K[t]_{0}^{t}\\ ln\left ( \frac{T_{2}-T_{0}}{T_{1}-T_{0}} \right )=-Kt

where K is a constant.

We have been given that the body cools from 80 o C to 50 o C in 5 minutes when the surrounding temperature is 20 o C.

T 2 = 50 o C

T 1 = 80 o C

T 0 = 20 o C

t = 5 min = 300 s.

\\ln\left ( \frac{50-20}{80-20} \right )=-K\times 300\\ K=\frac{ln(2)}{300}\\

For T 1 = 60 o C and T 2 = 30 o C we have

\\ln\left ( \frac{30-20}{60-20} \right )=-\frac{ln2}{300}t\\ t=\frac{ln(4)\times 300}{ln(2)} \\t=\frac{2ln(2)\times 300}{ln(2)}\\ t=600\ s\\ t= 10\ min

The body will take 10 minutes to cool from 60 o C to 30 o C at the surrounding temperature of 20 o C.

NCERT solutions for class 11 physics chapter wise

NCERT solutions for class 11 Subject wise

Importance of NCERT solutions for class 11 physics chapter 11 thermal properties of matter:

Thermal properties of matter is an important chapter for competitive exams. Questions based on calorimetry heat transfer and thermal expansions are frequently asked in competitive exams like NEET and JEE Mains exam. The NCERT solutions for class 11 are helpful to understand these concepts and their application in numerical problems.

Ncert Solutions for Class 11 Physics Thermal Properties of Matter

Source: https://school.careers360.com/ncert/ncert-solutions-class-11-physics-chapter-11-thermal-properties-of-matter