Zeroth and First law of Thermodynamics Problems With Solution

first law of thermodynamics problems and solutions

Zeroth law of Thermodynamics

The zeroth law of thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other.

First law of Thermodynamics

The first law of thermodynamics states that heat is a form of energy, and therefore thermodynamic processes are subject to the principle of conservation of energy. This means that heat energy can neither be created nor destroyed. However, it can be transferred from one place to another and converted into other forms of energy.

Problems With Solution

A  lead bullet Moving with a velocity of  40 ms-1 Falls down after striking a target. if only half of the heat produced is absorbed by the bullet,Find rise in its temperature.Specific heat of lead is 120 Jkg-1℃-1

Solution

let the mass of the bullet be m kg and its velocity be v.Then before striking the target the kinetic energy of the bullet = \frac {1} {2 } m v^2 joule.

on Striking the target this is converted into heat

∵ Q=\frac {1} {2 } m v^2 \;joule

As the bullet absorbs only half of the heat produced in the impact, heat received by the bullet 

Q’=0.5\; , \;Q= \frac {1} {2 } \times \frac {1} {2 } m v^2 \;joule

If the increase in the temperature of the bullet is ∆T℃,then

Q’=mc∆T  \; cal.

Hence, 

mc∆T=\frac {1} {2 } \times \frac {1} {2 } m v^2

∵ ∆T= \frac { v^2} {4c }

Given : v=40ms-1 , c=120 Jkg-1℃-1

{\color{DarkBlue} \Delta T= \frac { 40^2} {4 \times 120 }=3.33℃}

find the difference in the temperature of water at the top and bottom of a 420 m high waterfall if the value of j is 4.2 joule/cal.

Solution If a mass m kg of water falls from a height h m.

At the top of the fall potential energy = mgh Joule

At the bottom of the fall potential energy =0 

∵ Change In potential energy

\Delta U \;  = mgh\; joule

Change in potential energy is converted into the change in kinetic energy which in turn is converted into heat as water strikes ground.

∵ Increase in kinetic energy

\Delta K = mgh\; joule

∵ Heat produced

Q = \frac { \Delta K} {J} =\frac { mgh } { J }

If increase in the temperature of water is

 ΔT\; then \;Q=mcΔT

Where C is the specific heat of the water.

Thus,

mc \Delta T=\frac { mgh } { J }

∵ ΔT=gh/cJ

ΔT=\frac { gh } {cJ }

{\color{DarkBlue} Given :h=420m \; , g=9.8 \; ms^{-2} \; , c=1.0 \times 10^3 cal / kg \times \degree C,J =4.2J/cal.}

{ \color{DarkBlue} \Delta T=\frac { 9.8 \times  420}{1.0 \times 10^3 \times 4 . 2 } = 0.98 \degree C}

The height of a waterfall 50 m.Find the difference in the temperature of water at the top and bottom of the Fall (J=4.2 \times 10^{-1}erg/cal )

Solution

 as in above question, the change in temperature of water is given by

ΔT=\frac { gh } {cJ }

{\color{DarkBlue} Given :h=50m \; , g=981 \; cms^{-2} \; , c=1.0 \; cal / g \degree C , J =4.2 \times 10^7 erg/cal.}

{ \color{DarkBlue} \Delta T=\frac { 981 \times  5000}{4.2 \times 10^{-1} \times 1 } = 0.117\degree C}

A moving Bullet at 27℃ Strikes a obstacle and stops suddenly.The heat produced is received by the bullet and is  Just sufficient to melt it.Find the velocity of the bullet.

Given :Melting point of lead =327℃ , Specific heat of the lead =0.03,Latent heat of the lead =6 cal/g and J=4.2 joule/cal.

Solution

Let the mass and velocity of the bullet be m and v Respectively 

Its kinetic energy 

K= \frac {1} {2 } m v^2

Heat required to raise the temperature of the bullet to its melting point 

Q_1=mc \Delta T

Heat required to melt the bullet

Q_2=mL

Total heat required 

mc \Delta T+mL

Therefore from  W=JQ, we have

\frac {1} {2 } m v^2=J(m c \Delta T+m L)

v^2=2J(c \Delta T+L)

{ \color {Red}Given : J=4.2J/cal=4.2 \times 10^7 erg/cal , c=0.03 \; cal/g℃ \; , L=6 cal/g , ∆T= 327-27=300℃ }

v^2=2 \times 4.2 \times 10^7 \times (0.03 \times 300 +6)

=2 \times 4.2 \times 10^7 \times 15 =1.26 \times 10^9(cm/sec)^2

{ \color {DarkBlue} v=\sqrt {1.26 \times 10 ^ 9} = 3.5 \times 10^4 \; cm/s =350 \; m/s}

Exercise

1. If raindrops falls from a height of 10 km what is increase in the temperature (ans 23.45℃) Hint mc \Delta T =\frac{mgh}{J} ∵ \Delta T= \frac{gh}{cJ} ∵ \Delta T=\frac{9.8 \times 10 \times 10^3}{1 \times 10^3 \times 4.8}=23.45
2. The latent heat of stream is 536 cal g-1.Find its value in J kg-1 (ans 2.2512 ✖10^6 J kg-1)
3. What will be the rise in temperature of an iron ball falling from 25 m height, if Specific heat of iron = 0.1 cal.g^-1℃-1 and J=4.2 J cal-1? (ans 0.58℃)
4. Two Bullets having masses 10 gram and 15 gram Respectively strike a target with the equal velocity.If the entire heat produced in the Collision is used up in increasing the temperature of the bullets,For which bullet the rise in temperature will be more? (ans Increase of temperature will be same for both) Hint Q=\frac{W}{J} \; or\; mc \Delta T= \frac{\frac{1}{2}mv^2}{J}  

  \because \; \; \; \; \Delta T=\frac {v^2} {2 cJ} which \; is \; Independent \; of \; the \; mass \; of \; the \; bullet

• A metal ball falls from 10 M height and rebounds to 0.5 m Height. calculate rise in its temperature. specific heat of the metal of the Ball = 0.12 cal/g℃ (ans 0.18℃)
• A solid of 2 kg mass falls from a height of 3 km if the entire energy is converted into heat calculate the amount of heat produced (ans 14 \times 10^3 cal)
• From What height should a piece of ice fall so that it melts on striking the ground? ice absorbs half the amount of heat produced in the impact. (6700 km)
• Lead Bullet melts on striking a target assume that 25% of the heat produced is observed by the target.if the initial temperature of the bullet is 27℃,Find its velocity. M.P. \; of \;lead \;  327 \degree C,\; Specific\; heat \; of \;the\; lead=0.03 \; cal \; g^{-1} \degree C^{-1},Latent \; heat \; of \; melting \; of \; lead \; 6\\;;  cal /g^{-1} \; and J= 4.2 Jcal^{-1} 
• Height of a Niagra fall is 50 m find the difference in temperature at the top and bottom of the fall. given \; J=4.2\; J cal^{-1} \; , \; g=9.8 \; ms^{-2} ( ans \; 0.116 ℃)

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