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Q.1: Water flowing at the rate of 4.0 litres/minute is heated from 28 °C to 76 °C by a gas burner, if its heat of combustion is 4.0 x 10 4 Jig. Find the rate of consumption of fuel. Answer: Given: The rate of flow of water= 4.0 litres/min And, Heat of consumption = 4.0 x 1 o 4 Jig Since, the temperature of water is raised from 28 ·c to 76 ·c by the geyser Therefore, the initial temperature, T 1 = 28 °c And, the fi nal temperature, T 2 = 76 °c Hence, the r ise in temperature, 6T = T 2 - T 1 Therefore, 6T = 76 - 28= 48 °C And, specifi c heat of water, c = 4.2 J g - 1 0c - 1 Now, Mass of flowi ng water, m = 4.0 litres/min i.e. m = 4000 g/min [Since, 1L= 1000 g) Since, total heat used, 6Q = m x c x 6 T Therefore, 6Q= 4000 x 4.2 x 48= 8.064 x 10 5 Jlmin Therefore, the Rate of consumption of fuel= 4.0xlO Q.2: Find the amount of heat that must be provided to 1.8 x 10 - 2 kg of nitrogen at room temperature to increase its temperature at a constant pressure by 50 °C? ( Molecular mass of N 2 = 28 and take R = 8.3 J mol - 1 K - 1 ) Answer: Given: The Molecular mass of Nitrogen, M = 28 Universal gas Constant, R = 8.3 J mol - 1 K - 1 Mass of Nitrogen, m = 1.8 x 10 - 2 kg= 18 g _ s of N2 Now , the nu m ber o f m oles , n - Molicar ss of N Therefore, n = 0.643 m 18 M 28 Now, for Nitrogen Molar specific heat at constant pressure Cp = � x R = � x 8.3 Therefore, Cp = 29.05 J mol - 1 K - 1 Now, the total amount of heat that is to be supplied to i ncrease i ts temperature by 50 ·c: 6Q = n x Cp x 6 T 6Q = 0.643 X 29.05 X 50 = 933.9575J Therefore, the total amount of heat that is to be supplied to raise the temperature of Nitrogen by 50 0 c = 933.9575 J Q.3: Explain the following statements: (i). It is not necessary for two bodies having different temperatures say T 1 and T 2 , when brought in thermal contact to settle for a mean temperature of (T 1 + T 2 )/2 (ii). The coolant that is used in a chemical or nuclear power plant should have a high specific heat. (iii). While dri vi ng a vehicle, the air pressure of the tyres increases. (i). A harbour town has a more temperate climate than a desert town which is at the same latitude the NCERT SOLUTIONS CLASS-XI PHYSICS CHAPTER-12 THERMDYNAMICS

NCERT SOLUTIONS CLASS-XI PHYSICS CHAPTER-12 … · Mass of Nitrogen, m = 1.8 x 10 -2 kg= 18 g _ mass of N2 Now, the number of moles, n - Molicular mass of N Therefore, n = 0.643 m

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Page 1: NCERT SOLUTIONS CLASS-XI PHYSICS CHAPTER-12 … · Mass of Nitrogen, m = 1.8 x 10 -2 kg= 18 g _ mass of N2 Now, the number of moles, n - Molicular mass of N Therefore, n = 0.643 m

Q.1: Water flowing at the rate of 4.0 litres/minute is heated from 28 °C to 76 °C by a gas burner, if its

heat of combustion is 4.0 x 104 Jig. Find the rate of consumption of fuel.

Answer:

Given:

The rate of flow of water= 4.0 litres/min

And, Heat of consumption = 4.0 x 1 o4 Jig

Since, the temperature of water is raised from 28 ·c to 76 ·c by the geyser

Therefore, the initial temperature, T1 = 28 °c

And, the final temperature, T2 = 76 °c

Hence, the rise in temperature, 6. T = T 2 - T 1

Therefore, 6. T = 76 - 28 = 48 °C

And, specific heat of water, c = 4.2 J g -1 0c - 1

Now, Mass of flowing water, m = 4.0 litres/min

i.e. m = 4000 g/min [Since, 1L = 1000 g)

Since, total heat used, 6.Q = m x c x 6. T

Therefore, 6.Q = 4000 x 4.2 x 48 = 8.064 x 105 Jlmin

Therefore, the Rate of consumption of fuel=

s.oMxt�• = 2.016 g/min

4.0xlO

Q.2: Find the amount of heat that must be provided to 1.8 x 1 0 - 2 kg of nitrogen at room temperature to increase its temperature at a constant pressure by 50 °C? ( Molecular mass of N2 = 28 and take R =

8.3 J mol - 1 K - 1 )

Answer: Given: The Molecular mass of Nitrogen, M = 28

Universal gas Constant, R = 8.3 J mol -1 K -1

Mass of Nitrogen, m = 1.8 x 10 -2 kg= 18 g _ mass of N2 Now, the number of moles, n - Molicular mass of N

Therefore, n = 0.643

m 18

M 28

Now, for Nitrogen Molar specific heat at constant pressure Cp = � x R = � x 8.3

Therefore, Cp = 29.05 J mol -1 K -1

Now, the total amount of heat that is to be supplied to increase its temperature by 50 ·c:

6.Q = n x Cp x 6. T ⇒ 6.Q = 0.643 X 29.05 X 50 = 933.9575J Therefore, the total amount of heat that is to be supplied to raise the temperature of Nitrogen by 50 0c = 933.9575 J

Q.3: Explain the following statements:

(i). It is not necessary for two bodies having different temperatures say T1 and T2, when brought in thermal contact to settle for a mean temperature of (T1 + T2)/2

(ii). The coolant that is used in a chemical or nuclear power plant should have a high specific heat.

(iii). While driving a vehicle, the air pressure of the tyres increases.

(i). A harbour town has a more temperate climate than a desert town which is at the same latitude the

NCERT SOLUTIONS CLASS-XI PHYSICS

CHAPTER-12 THERMDYNAMICS

Page 2: NCERT SOLUTIONS CLASS-XI PHYSICS CHAPTER-12 … · Mass of Nitrogen, m = 1.8 x 10 -2 kg= 18 g _ mass of N2 Now, the number of moles, n - Molicular mass of N Therefore, n = 0.643 m
Page 3: NCERT SOLUTIONS CLASS-XI PHYSICS CHAPTER-12 … · Mass of Nitrogen, m = 1.8 x 10 -2 kg= 18 g _ mass of N2 Now, the number of moles, n - Molicular mass of N Therefore, n = 0.643 m
Page 4: NCERT SOLUTIONS CLASS-XI PHYSICS CHAPTER-12 … · Mass of Nitrogen, m = 1.8 x 10 -2 kg= 18 g _ mass of N2 Now, the number of moles, n - Molicular mass of N Therefore, n = 0.643 m
Page 5: NCERT SOLUTIONS CLASS-XI PHYSICS CHAPTER-12 … · Mass of Nitrogen, m = 1.8 x 10 -2 kg= 18 g _ mass of N2 Now, the number of moles, n - Molicular mass of N Therefore, n = 0.643 m