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

Problem 1:-A spherical constant temperature heat source of radius r

_{1}is at the center of a uniform solid sphere of radius r_{2}. Find out the rate which is proportional to heat transferred through the surface of the sphere.

Solution:-The rate

Hat 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 Δ

x.(c) directly proportional to the temperature difference Δ

T.So,

H=kAΔT/ ΔxWhere

kis the proportionality constant and is called thermal conductivity of the material.From above we know that, the rate

Hat which heat is transferred through the slab is directly proportional to the area (A) available.Area

Aof solid sphere is defined as,

A= 4πr^{2}Here

ris the radius of sphere.So, the area

A_{1}of uniform small solid sphere having radiusr_{1}will be,

A_{1}= 4πr_{1}^{2}And, the area

A_{2}of uniform large solid sphere having radiusr_{2}will be,

A_{2}= 4πr_{2}^{2}Thus the area

Afrom which heat is transferred through the surface of the sphere will be the difference of area of uniform large solid sphereA_{2}and small solid sphereA_{1}.So,

A=A_{2}-A_{1}= 4π

r_{2}^{2}- 4πr_{1}^{2}= 4π (

r_{2}^{2}-r_{1}^{2})Since the rate

Hat 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 tor_{2}^{2}-r_{1}^{2}.

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/mΔTSo,

Q=c mΔTHere, the heat transferred is

Q, specific heat capacity isc, mass ismand 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

Lfor the process. The total heatQtransferred in a phase change is then,

Q=LmHere

mis the mass of the sample that changes phase.

Solution:-The heat given off the steam

Q_{s}will be equal to,

Q_{s}=m_{s}L_{v}+m_{s}c_{w}ΔTHere, mass of steam is

m_{s}, latent heat vaporization isL_{v}, specific heat capacity of water isc_{w}and the temperature difference is ΔT.The heat taken in by the ice

Q_{i}will be equal to,

Q_{i}=m_{i}L_{f}+m_{i}c_{w}ΔTHere, mass of ice is

m_{i}, latent heat fusion isL_{f}, specific heat capacity of water isc_{w}and the temperature difference is ΔT.Heat given off the steam

Q_{s}is equal to the heat taken in by the iceQ_{i}.So,

Q_{s}=Q_{i}

m_{s}L_{v}+m_{s}c_{w}ΔT=m_{i}L_{f}+m_{i}c_{w}ΔT

m_{s}(L_{v}+c_{w}ΔT) =m_{i}(L_{f}+c_{w}ΔT)

m_{s}=m_{i}(L_{f}+c_{w}ΔT)/ (L_{v}+c_{w}ΔT)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 icem_{i}, 333×10^{3}J/kg forL_{f}, 4190 J/kg.K forc_{w}, 50^{°}C for ΔT, 2256×10^{3}J/kg forL_{v}in the equationm_{s}=m_{i}(L_{f}+c_{w}ΔT)/ (L_{v}+c_{w}ΔT), we get,

m_{s}=m_{i}(L_{f}+c_{w}ΔT)/ (L_{v}+c_{w}ΔT)=(150 g)[(333×10

^{3}J/kg) +(4190 J/kg.K) (50^{°}C)]/ [(2256×10^{3}J/kg)+ (4190 J/kg.K) (50^{°}C)]=(150 g×(10

^{-3}kg/1g))[(333×10^{3}J/kg) +(4190 J/kg.K) (50+273)K]/ [(2256×10^{3}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

Wis defined as,

W=mgΔyHere

mis the mass,gis the free fall acceleration and Δyis the increase in height.The specific heat capacity

cof a material is equal to the heat capacityCper unit massmof the body.So,

c=C/m=

Q/mΔT(Since,C=Q/ΔT)Here Δ

Tis the increase in temperature.So,

Q=mcΔTAs, |

Q|= |W|,

mcΔT=mgΔySo, Δ

T=gΔy/c

Solution:-(a) To obtain the possible increase in temperature Δ

T, substitute 9.81 m/s^{2}forg, 49.4 m for Δyand 4190 J/kg.K for specific heat capacity of watercin the equation ΔT=gΔy/c,Δ

T=gΔy/c= (9.81 m/s

^{2}) (49.4 m)/ (4190 J/kg.K)= (9.81 m/s

^{2}) (49.4 m)/ (4190 J/kg.K) (1 kg.m^{2}/s^{2}/1 J)= 0.116 K

From the above observation we conclude that, the possible increase in temperature Δ

Twould 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

Qthat must be given to a body of massm, whose material has a specific heatc, to increase its temperature from initial temperatureT_{i}to final temperatureT_{f}is,

Q=mc(T_{f}-T_{i})So,

m=Q/ c(T_{f}-T_{i})Density

ρis equal to massmper unit volumeV.So,

ρ=m/VSo volume

Vwill 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

mof water, substitute 5.22 GJ forQ, 4190 J/kg. K for specific heat capacitycof water, 50.0^{°}C forT_{f}and 22.0 ° C forT_{i}in the equationm=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) (10

^{9}J/1 GJ)/(4190 J/kg. K) ((50.0+273) K –(22.0+273) K)= (5.22 ×10

^{9}J)/(4190 J/kg. K) (28 K)= 4.45×10

^{4}kgTo obtain the volume

Vof water, substitute 4.45×10^{4}kg for massmand 998 kg/m^{3}for densityρof water in the equationV=m/ρ,

V=m/ρ= (4.45×10

^{4}kg) / (998 kg/m^{3})= 44.5 m

^{3}From the above observation we conclude that, the volume

Vof water will be 44.5 m^{3}.

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

Qthat must be given to a body of massm, whose material has a specific heatc, to increase its temperature from initial temperatureT_{i}to final temperatureT_{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

Solution:-To obtain the time required to bring this water from 23.5

^{°}C to the boiling point, substitute 136 g for mass of waterm, 4190 J/kg. K for specific heat capacity of waterc, 100^{°}C for final temperatureT_{f}(boiling point of water), 23.5° C for initial temperatureT_{i}and 220 watts for powerPin the equationt=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.

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