Solved Problems on Specific Heat, Latent Heat and Entropy:- 

Problem 1:-

A spherical constant temperature heat source of radius r1 is at the center of a uniform solid sphere of radius r2. Find out the rate which is proportional to heat transferred through the surface of the sphere.

Solution:-

The rate H at which heat is transferred through the slab is,

(a) directly proportional to the area (A) available.

(b) inversely proportional to the thickness of the slab Δx.

(c) directly proportional to the temperature difference ΔT.

So, H = kA ΔT/ Δx    

Where k is the proportionality constant and is called thermal conductivity of the material.

From above we know that, the rate H at which heat is transferred through the slab is directly proportional to the area (A) available.

Area A of solid sphere is defined as,

A = 4πr2

Here r is the radius of sphere.

So, the area A1 of uniform small solid sphere having radius r1 will be,

A1 = 4πr12

And, the area A2 of uniform large solid sphere having radius r2 will be,

A2 = 4πr22

Thus the area A from which heat is transferred through the surface of the sphere will be the difference of area of uniform large solid sphere A2 and small solid sphere A1.

So, A = A2 - A1

          = 4πr22 - 4πr12

         = 4π (r22 - r12)

Since the rate H at which heat is transferred through the slab is directly proportional to the area (A) available, therefore the rate at which heat is transferred through the surface of the sphere is proportional to r22 - r12.

Problem 2:-

What mass of steam at 100°C must be mixed with 150 g of ice at 0°C, in a thermally insulated container, to produce liquid water at 50°C. 

Concept:-

The heat capacity per unit mass of a body, called specific heat capacity or usually just specific heat, is characteristic of the material of which the body is composed.

c = C/m

  = Q/mΔT

So, Q = c mΔT

Here, the heat transferred is Q, specific heat capacity is c, mass is m and the temperature difference is ΔT.

The amount of heat per unit mass that must be transferred to produce a phase change is called the heat of transformation or latent heat L for the process. The total heat Q transferred in a phase change is then,

Q = Lm

Here m is the mass of the sample that changes phase.

Solution:-

The heat given off the steam Qs will be equal to,

Qs = msLv+ mscwΔT

Here, mass of steam is ms, latent heat vaporization is Lv, specific heat capacity of water is cw and the temperature difference is ΔT.

The heat taken in by the ice Qi will be equal to,

Qi = miLf+ micwΔT

Here, mass of ice is mi, latent heat fusion is Lf, specific heat capacity of water is cw and the temperature difference is ΔT.

Heat given off the steam Qs is equal to the heat taken in by the ice Qi.

So, Qs = Qi

msLv+ mscwΔT = miLf+ micwΔT

ms(Lv+ cwΔT) = mi(Lf+ cwΔT)

ms = mi(Lf+ cwΔT)/ (Lv+ cwΔT)

To obtain the mass of the steam at 100 ̊C must be mixed with 150 g of ice at 0 ̊C, substitute 150 g for mass of ice mi, 333×103 J/kg for Lf, 4190 J/kg.K for cw, 50̊ C for ΔT, 2256×103 J/kg for Lv in the equation ms = mi(Lf+ cwΔT)/ (Lv+ cwΔT), we get,

ms = mi(Lf+ cwΔT)/ (Lv+ cwΔT)

=(150 g)[(333×103 J/kg) +(4190 J/kg.K) (50̊ C)]/ [(2256×103 J/kg)+ (4190 J/kg.K) (50̊ C)]

 =(150 g×(10-3 kg/1g))[(333×103 J/kg) +(4190 J/kg.K) (50+273)K]/ [(2256×103 J/kg)+ (4190 J/kg.K) (50+273)K]

= 0.033 kg

From the above observation we conclude that, the mass of steam at 100 ̊C must be mixed with 150 g of ice at 0 ̊C would be 0.033 kg.

Problem 3:-

(a) Compute the possible increase in temperature for water going over Niagara Falls, 49.4 m high. (b) What factors would tend to prevent this possible rise?

Concept:-

Work done W is defined as,

W = mgΔy

Here m is the mass, g is the free fall acceleration and Δy is the increase in height.

The specific heat capacity c of a material is equal to the heat capacity C per unit mass m of the body.

So, c = C/m

          = Q/mΔT     (Since, C = QT )

Here ΔT is the increase in temperature.

So, Q = mcΔT

As, |Q| = |W|,

mcΔT = mgΔy

So, ΔT = gΔy/c

Solution:-

(a) To obtain the possible increase in temperature ΔT, substitute 9.81 m/s2 for g, 49.4 m for Δy and 4190 J/kg.K for specific heat capacity of water c in the equation ΔT = gΔy/c,

ΔT = gΔy/c

     = (9.81 m/s2) (49.4 m)/ (4190 J/kg.K)

     = (9.81 m/s2) (49.4 m)/ (4190 J/kg.K) (1 kg.m2/s2 /1 J)

     = 0.116 K

From the above observation we conclude that, the possible increase in temperature ΔT would be 0.116 K

(b) The above expression is valid only there is no loss energy. But due to the viscosity of the water, some of the energy is lost in rising against the gravity. So, the factor which prevents the water from rising is the viscosity.

Problem 4:-

In a certain solar house, energy from the Sun is stored in barrels filled with water. In a particular winter stretch of five cloudy days, 5.22 GJ are needed to maintain the inside of the house at 22.0°C. Assuming that the water in the barrels is at 50.0°C, what volume of water is required?  

Concept:-

Heat Q that must be given to a body of mass m, whose material has a specific heat c, to increase its temperature from initial temperature Ti to final temperature Tf is,

Q = mc (Tf - Ti)

So, m = Q/ c (Tf - Ti)

Density ρ is equal to mass m per unit volume V.

So, ρ = m/V

So volume V will be,

V = m/ρ

Solution:-

To find the volume water, first we have to find out the mass of water which is required to transfer 5.22 GJ amount of heat energy.

To find the mass m of water, substitute 5.22 GJ for Q, 4190 J/kg. K for specific heat capacity c of water, 50.0 ̊ C for Tf and 22.0 ̊ C for Ti in the equation m = Q/ c (Tf - Ti),

m = Q/ c (Tf - Ti)

    = 5.22 GJ/(4190 J/kg. K) (50.0 ̊ C-22.0 ̊ C)

    = (5.22 GJ) (109 J/1 GJ)/(4190 J/kg. K) ((50.0+273) K –(22.0+273) K)

  = (5.22 ×109 J)/(4190 J/kg. K) (28 K)

  = 4.45×104 kg

To obtain the volume V of water, substitute 4.45×104 kg for mass m and 998 kg/m3 for density ρ of water in the equation V = m/ρ,

V = m/ρ

   = (4.45×104 kg) / (998 kg/m3)

   = 44.5 m3

From the above observation we conclude that, the volume V of water will be 44.5 m3.

Problem 5:-

A small electric immersion heater is used to boil 136 g of water for a cup of instant coffee. The heater is labeled 220 watts. Calculate the time required to bring this water from 23.5°C to the boiling point, ignoring any heat losses. 

Concept:-

Heat Q that must be given to a body of mass m, whose material has a specific heat c, to increase its temperature from initial temperature Ti to final temperature Tf is,

Q = mc (Tf - Ti)

But time (t) is equal to the heat energy (Q) divided by power (P).

t = Q/ P

 = mc (Tf - Ti)/P

Solution:-

To obtain the time required to bring this water from 23.5̊ C to the boiling point, substitute 136 g for mass of water m, 4190 J/kg. K for specific heat capacity of water c, 100̊ C for final temperature Tf (boiling point of water), 23.5̊ C for initial temperature Ti and 220 watts for power P in the equation t = mc (Tf - Ti)/P,

t = mc (Tf - Ti)/P

  = (136 g) (4190 J/kg. K) (100̊ C - 23.5̊ C) / 220 W

 = (136 g×10-3 kg/1 g) (4190 J/kg. K) (100̊ C - 23.5̊ C) / 220 W

= (0.136 kg) (4190 J/kg. K) ((100+273) K – (23.5 + 273)K) / 220 W

= (0.136 kg) (4190 J/kg. K) (373 K – 296.5 K) / 220 W

= (0.136 kg) (4190 J/kg. K) (76.5 K) / 220 W

= 198.15 s

Rounding off to three significant figures, the time required to bring this water from 23.5̊ C to the boiling point would be 198 s.

 

Related Resources: You might like to refer some of the related resources listed below:

Click here for the Detailed Syllabus of IIT JEE Physics.

Look into the Sample Papers of Previous Years to get a hint of the kinds of questions asked in the exam.

 

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