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.

Area A of solid sphere is defined as,

A = 4πr²

Here r is the radius of sphere.

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.

Q = Lm

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

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

Q_s = m_sL_v+ m_sc_wΔT

Here, mass of steam is m_s, latent heat vaporization is L_v, specific heat capacity of water is c_w and the temperature difference is ΔT.

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

Q_i = m_iL_f+ m_ic_wΔT

Here, mass of ice is m_i, latent heat fusion is L_f, specific heat capacity of water is c_w and the temperature difference is ΔT.

Heat given off the steam Q_s is equal to the heat taken in by the ice Q_i.

So, Q_s = Q_i

m_sL_v+ m_sc_wΔT = m_iL_f+ m_ic_wΔT

m_s(L_v+ c_wΔT) = m_i(L_f+ c_wΔT)

m_s = m_i(L_f+ c_wΔT)/ (L_v+ c_wΔT)

m_s = m_i(L_f+ c_wΔT)/ (L_v+ c_wΔT)

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

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

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 T_i to final temperature T_f is,

Q = mc (T_f - T_i)

So, m = Q/ c (T_f - T_i)

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

So, ρ = m/V

So volume V will be,

V = m/ρ

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 T_f and 22.0 ° C for T_i in the equation m = Q/ c (T_f - T_i),

m = Q/ c (T_f - T_i)

    = 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

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 T_i to final temperature T_f is,

Q = mc (T_f - T_i)

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

t = Q/ P

 = mc (T_f - T_i)/P

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

t = mc (T_f - T_i)/P

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

 = (136 g×10^-3 kg/1 g) (4190 J/kg. K) (100^° C - 23.5° 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 23.5^° C to the boiling point would be 198 s.

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

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