Thermodynamics Numerical on Heat and Work for Class-11 ISC Physics Nootan Solutions Ch-19. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

## Thermodynamics Numerical on Heat and Work for Class-11 ISC Physics Nootan Solutions

Board | ISC |

Class | 11 |

Subject | Physics |

Book | Nootan |

Chapter-19 | Thermodynamics |

Topics | Numerical on Heat and Work |

Academic Session | 2024-2025 |

### Numerical on Heat and Work

(Thermodynamics Numerical for Class-11 ISC Physics Nootan Solutions Ch-19)

**Que-1: How much maximum work can be obtained from 100 calories of heat?**

**Ans: **Joule = Calorie x 4.18

=> 100 x 4.18 = **418 J**

**Que-2: Latent heat of steam is 536 Cal g^-1. Express it in J kg^-1.**

**Ans: **536 Cal g^-1

=> 536 x 418 J x (10^-3)^-1 Kg

=> **2.24 x 10^6 J Kg^-1**

**Que-3: Specific heat of water is 1 Cal g^-1 °C^-1. Express it in J kg^-1 °C^-1.**

**Ans: **1 Cal g^-1 °C^-1

=> 4.18 J x (10^-3)^-1

=> **4.18 x 10³ J Kg^-1 °C^-1**

**Que-4: Specific heat of copper is 0.092 Cal g^-1°C^-1. Express it in J kg^-1 K^-1.**

**Ans: **0.092 Cal g^-1 °C^-1

=> 0.92 x 4.18 J (10^-3 Kg)^-1 K^-1

=> **385 J Kg^-1 K^-1**

**Que-5: If in stopping a car of mass 800 kg by applying brakes, 41.8 kcal of heat is produced, then what was the speed of the car before applying the brakes? **

**Ans: **K = 1/2 m v² = 1/2 x 800 x v² = 41.8 x 4.18 x 10³

=> 400 v² = 41.8 x 4.18 x 10^4

=> 20 v = 4.18 x 10²

=> v = 418/20

=> v = **20.9 m s^-1 **

**Que-6: A bullet of mass 20 g strikes a target with a velocity 100 m s^-1 and comes to rest. 50% kinetic energy of the bullet raises its temperature by 25°C. Calculate : (i) increase in the internal energy of the bullet due to rise in temperature, (ii) specific heat of the material of the bullet. **

**Ans: (i) **Increase in internal energy

=> 1/2 x 1/2 m v²

=> 1/2 x 1/2 x 0.020 x (100)²

=> **50 J**

**(ii) **again, Q = m c Δt

=> 50 = 0.02 x c x 25

=> c = 50 / (0.02 x 25)

=> c = **100 J Kg^-1 °C^-1**

**Que-7: A lead bullet of specific heat 0.032 kcal kg^-1 °C^-1 strikes a target with a velocity of 300 m s^-1. If the bullet is completely stopped by the target, find the rise in the temperature of the bullet. Assume that the heat produced is equally shared by the target and the bullet. (J = 4.2 J Cal^-1) **

**Ans: **Energy of bullet is equally shared with target

=> 1/2 x 1/2 m v² = m c Δt

=> 1/4 x m x (300)² = m x 0.032 x 4.20 x 10³ x Δt

=> Δt = (300 x 300) / (4 x 0.032 x 4.20 x 10³)

=> Δt = **167.4 °C**

**Que-8: A bullet of lead just melts when stopped by an obstacle. Assuming that 25% of the heat produced is absorbed by the obstacle, find the velocity of the bullet if its initial temperature is 27°C. J = 4.2 J Cal^-1, melting point of lead = 327 °C, specific heat of lead = 0.03 Cal g^-1 °C^-1, latent heat of melting of lead = 6 Cal g^-1,**

**Ans: **75/100 x 1/2 x m v² = m c Δt

=> 75/100 x 1/2 x v² = 0.03 x 4.2 x 10³ x (327 – 273) + 6 x 4.2 x 10³

=> v = **410 m s^-1**

**Que-9: A 2 kg sphere falls from a height of 3 m. If whole of its potential energy is converted into heat, then how much heat will be produced? (1 Cal = 4.2 J).**

**Ans: ** Energy converted into heat

=> m g h/4.2 Cal= 2 x 9.8 x 3 / 4.2 = **14 Cal**

**Q-10: A 5 kg hammer falls with a velocity 1 m s^-1 on a piece of lead of mass 300 g at 27°C. How many strokes will be required for melting the lead?**

**Ans: **Let it needed n strikes then

=> m x 1/2 m v² = M c Δt + M L {M = mass of lead piece}

=> n x 1/2 x 5 x(1)² = 0.3 (125 x 300 + 6 x 4.2 x 10³)

=> n = **7524**

**Que-11: A lead ball of mass 2.0 kg falls from a height of 30.0 m. If the total kinetic energy of the ball is converted into heat and remains in it, calculate the rise in temperature of the ball when it strikes the ground. Find the change in the internal energy of the ball in this process. (Specific heat of lead = 125 J kg^-1 °C^-1)**

**Ans: **m g h = m c Δt

=> 9.8 x 30 = 125 x Δt

=> Δt = **2.35 °C**

again increase in energy

=> m g h = 2 x 9.8 x 30 = **588 J**

**Que-12: Water falls from a height of 20 m at a rate of 100 kg s^-1. How many calorie of heat will be produced per second on striking with the earth? Assume that the whole energy is converted into heat.**

**Ans: **Heat produced per sec

=> m g h / 4.2

=> 100 x 7 x 20 / 3

=> 14000 / 3

=> 4.7 x 10³ Cal = **4.7 kcal**

**Que-13: From what height should a piece of ice fall upon the ground so that it is completely melted, assuming that all the heat produced remains in ice? (Latent heat of ice= 80 Cal g^-1, J = 4.2 J Cal^-1, g = 9.8 N kg^-1.)**

**Ans: **m g h = m L

=> h = L/g

=> h = 80 x 4.2 x 10³ / 9.8

=> h = **34.28 km**

**Que-14: In a waterfall, water falls from 420 m on the ground. Determine the rise a height of in the temperature of water, if total heat generated remains in water. (g = 10 m s^-2 and J = 4.2 x 10^3 J kcal^-1) **

**Ans: **m g h = m c Δt

=> Δt = g h / c

=> 10 x 420 / 4.2 x 10³

=> **1 °C**

**Que-15: A 0.1 kg steel ball falls from a height of 10 m on the ground and rebounds to a height of 7 m. (i) Why does it not rebound to its original height? (ii) If the dissipated energy remains in the ball in the form of heat, then by how much will the temperature of the ball be raised?**

**Ans: (i) **Collision between the earth and the ball is non-elastic in which energy is dissipated.

**(ii)** m g h = m c Δt

=> Δt = g h / c

=> 9.8 x (10 – 7) / 0.11 x 4.2 x 10³

=> **0.064 °C**

**Que-16: What amount of work will have to be done in converting 1 kg of water into steam at 100°C and normal atmospheric pressure? The volume of 1 kg water at 100°C is 10^-3 m³ and that of 1 kg of steam is 10 1.671 m³ and the normal pressure is 10^5 N m^-2.**

**Ans: **ΔW = P x ΔV

=> 10^5 (1.671 – 0.001)

=> **1.670 x 10^5 J**

**Que-17: A gas expands from 75 litre to 125 litre at a constant pressure of 4 atmospheric. If one atmospheric pressure is 1.0 × 10^5 Pa (N m^-2), calculate the work done by the gas during this expansion.**

**Ans: **ΔW = P x ΔV

ΔW = 4 x 1 x 10^5 (125 – 75) x 10^-3

=> **2.0 x 10^4 J**

**Que-18: A 0.1 kg ball falls from a height of 10 m and rebounds to a height of 7.0 m. What is the change in the internal energy of the ball and the earth?**

**Ans: **Increase in energy of earth ball system = decrease in Potential energy

=> m g (10 – 7)

=> 0.1 x 9.8 x 3 = **2.9 J**

— : end of Thermodynamics Numerical on Heat and Work for Class-11 ISC Physics Nootan Solutions :–

Return to : **– Nootan Solutions for ISC Class-11 Physics**

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