Chapter One Basics of Heat Transfer

1.1- Thermodynamics and Heat Transfer

The science of thermodynamics deals with the amount of heat transfer as a system undergoes a process
from one equilibrium state to another, and makes no reference to how long the process will take. Whereas
the science of heat transfer deals with the rate of heat transfer, which is the main quantity of interest in
the design and evaluation of heat transfer equipment.

1.1.1- Heat and Other Forms of Energy

Energy can exist in numerous forms such as thermal, mechanical, kinetic,
potential, electrical, magnetic, chemical, and nuclear, and their sum constitutes the
total energy E or (e on a unit mass basis). The forms of energy related to the
molecular structure of a system and the degree of the molecular activity are referred
to as the microscopic energy. The sum of all microscopic forms of energy is called
the internal energy of system, and is denoted U (or u on a unit mass basis).
Internal energy may be viewed as the sum of kinetic and potential energies of the
molecules. The portion of the internal energy of a system associated with the kinetic
energy of the molecules is called sensible energy or sensible heat, where the
internal energy associated with the phase change of a system is called latent
energy or latent heat. Fig.(1.1) The internal
energy (u) represents
In the analysis of a system that involve fluid flow as shown in Fig.(1.1), a the microscopic energy
combination of properties u and Pv is obtained to defined by a property called of a non-flowing fluid,
enthalpy h, where; h=u+Pv and the term Pv represents the flow energy of the fluid whereas enthalpy (h)
(also called the flow work). represents the
microscopic energy of
a flowing- fluid

1.1.2- Specific Heats of Gases, Liquids, and Solids

The ideal gas is defined as a gas that obeys the following relation;
P v = RT , or; P =  RT (1.1)
where P is the absolute pressure, v is the specific volume, T is the absolute temperature, ρ is the density,
and R is the gas constant.
Note1- The specific heat is defined as the energy required to raise the temperature of a unit mass of a
substance by one degree. In general, this energy depends on how the process is executed. In
thermodynamics, we are interested in two kinds of specific heats; specific heat at constant volume Cv
and specific heat at constant pressure Cp. For ideal gases, these two specific heats are related to each
other by Cp= Cv + R, where the units of Cp, Cv, and R are [kJ/kg.K˚].
Note2- The change in internal energy and enthalpy can calculated for ideal gas by the following relations;
u = Cv T and h = C p T ( J / g ) (1.2.a)
or;
U = m Cv T and H = m C p T ( J ) (1.2.b)
where m is the mass of the system.
Note3- A substance whose specific volume (or density) does not change with temperature or pressure is
called an incompressible substance. The specific volumes of solids and liquids generally remain
constant during a process, and thus they can be approximated as incompressible substances, where the
constant-volume and constant-pressure specific heats are identical for incompressible substances.
Therefore, for solids and liquids we have; Cp≈Cv≈C, where C represents specific heat of the
incompressible substances. Therefore, the change in the internal energy of solids and liquids can be
expressed as;
U = m C T (J )  (1.3)

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Chapter One Basics of Heat Transfer
1.1.3- The First Law of Thermodynamics
The first law of thermodynamics, also known as the conservation of energy principle or (the
energy balance) which for any system undergoing any process may be expressed as follows: The net
change (increase or decrease) in the total energy of the system during a process is equal to the difference
between the total energy entering and the total energy leaving the system during that process, That is;
 Total energy   Total energy   Change in the total 
  −   =   (1.4)
 entering the system   leaving the system   energy of the system 

Noting that energy can be transferred to or from a system by heat, work, and
mass flow, and that the total energy of a simple compressible system consists
of internal, kinetic, and potential energies. So, the energy balance for any
system undergoing any process can be expressed as;
Ein − Eout = Esystem (J )  (1.5.a)
   
Net energytransfer Changein internal,kinetic,
by heat,work,and mass potential,etc., energies
Fig.(1.2) In steady operation,
or, in the rate form, as; the rate of energy transfer to
E in − E out = Esystem / dt (W )  (1.5.b) a system is equal to the rate
    of energy transfer from the
Rateof net energytransfer Rateof changein internal system
by heat,work,and mass kinetic,potential,etc., energies

Note- The energy change of a system is zero (ΔEsystem = 0) if the state of the
system does not change during the process, that is, the process is steady. So,
as shown in Fig.(1.2), the energy balance in this case reduces to;
E in = E out (W )  (1.6)
 
Rateof net energytransfer Rateof net energytransfer
by heat,work,and mass by heat,work,and mass

1.1.3.a- Energy Balance for Closed Systems (Fixed Mass)

For a closed system consists of a fixed mass, then the total energy E will
consist of the internal energy U. This is especially the case for stationary
systems since they don’t involve any changes in their velocity or elevation
during a process. So, the energy balance relation for stationary closed system
will be as follow;
Ein − Eout = U = m Cv T (J ) (1.7.a)
when the system involves heat transfer only and no work interactions across
Fig.(1.3) In the absence of
its boundary as shown in Fig.(1.3), the energy balance relation further reduces any work interactions, the
to; change in the internal energy
Q = m Cv T (J ) (1.7.b) of a closed system is equal to
the net heat transfer
where Q is the net amount of heat transfer to or from the system.

1.1.3.b- Energy Balance for Steady-Flow Systems

For a steady-flow system with one inlet and one exit, the rate of mass flow
into the control volume must be equal to the rate of mass flow out of it, thus,
(m in = m out = m ) . When the changes in kinetic and potential energies are
negligible, and there is no work interaction as shown in Fig.(1.4), the energy
balance for such a steady-flow system reduces to; Fig.(1.4) The net rate of energy
Q = m h = m C p T ( J / s = W ) (1.8) transfer to a fluid in a control
volume is equal to the rate of
where Q is the rate of net heat transfer into or out of the control volume. increase in the energy of the
fluid stream flowing through the
control volume

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Chapter One Basics of Heat Transfer
1.1.4- Surface Energy Balance
A surface contains no volume or mass, and thus no energy. Therefore, a
surface can be viewed as a fictitious system whose energy content remains
constant during a process (just like a steady-state or steady-flow system). Then
the energy balance for a surface can be expressed as;
E in = E out (W )  (1.9)
This relation is valid for both steady and transient conditions, and the surface
energy balance does not involve heat generation since a surface does not have a
volume. For example, the energy balance for the outer surface of the wall in
Fig.(1.5) can be expressed as; Q1 = Q2 + Q3
Fig.(1.5) Energy
1.1- interactions at the outer
wall surface of a house

as in Fig.(1.6).

Fig.(1.6) Schematic for

Example-1.1

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Chapter One Basics of Heat Transfer

1.2-

as in Fig.(1.7).

Fig.(1.7) Schematic for

Example-1.2

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Chapter One Basics of Heat Transfer

1.2- Heat Transfer Modes

The heat is defined as the form of energy that can be transferred from one system to another as a result
of temperature difference. The science that deals with the determination of the rates of such energy
transfers is the heat transfer. The transfer of energy as heat is always from the higher-temperature
medium to the lower-temperature one, and heat transfer stops when the two mediums reach the same
temperature. Heat can be transferred in three different modes: conduction, convection, and radiation. All
modes of heat transfer require the existence of a temperature difference, and all modes are from the high-
temperature medium to a lower-temperature one.
1.2.1- Conduction
Conduction is the transfer of energy from the more
energetic particles of a substance to the adjacent less energetic
ones as a result of interactions between the particles.
Conduction can take place in solids, liquids, or gases. In solids,
conduction is due to the combination of vibrations of the
molecules in a lattice and the energy transport by free
electrons, while in gases and liquids, it is due to the collisions
and diffusion of the molecules during their random motion, as
shown in Fig.(1.8).
The rate of heat conduction through a medium depends on the
geometry of the medium, its thickness, and the material of the
medium, as well as the temperature difference across the
medium.
Experiments have shown that the rate of heat transfer Q
through the wall will increase when the temperature difference
(ΔT) across the wall or the area (A) normal to the direction of
heat transfer increases, while it will decrease with the increase
of the wall thickness (L). Thus we conclude that the rate of
heat conduction through a plane layer is directly proportional
to the temperature difference across the layer ΔT and the heat
transfer area A, but is inversely proportional to the thickness
of the layer L. Thus;
( Area) (Temperature difference) Fig.(1.8) The mechanisms of heat conduction
Rate of heat conduction  in different phases of a substance
Thickness
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Chapter One Basics of Heat Transfer
If we consider steady heat conduction through a large plane wall
of thickness Δx = L and area A, as shown in Fig.(1.9), where the
temperature difference across the wall is ΔT = T2 – T1, so, the rate
of heat conduction will be;
• T −T T
Q Cond. = k A 1 2 = −k A (W ) (1.10)
x x
where the constant of proportionality k is the thermal
conductivity of the material, which is a measure of the ability of
a material to conduct heat. In the limiting case of Δx→0, the
equation above reduces to the differential form as follow;
• dT
Q Cond. = −k A (W ) (1.11)
dx
which is called Fourier’s law of heat conduction. Here dT/dx is
the temperature gradient, which is (the rate of change of T with
x) at location x.
Fig. (1.9) Heat conduction through a large
plane wall of thickness Δx and area A

1.3-

, Fig.(1.10).

Fig.(1.10) Schematic for

Example-1.3

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Chapter One Basics of Heat Transfer

1.2.1.a- Thermal Conductivity

The thermal conductivity of a material can be defined
as the rate of heat transfer through a unit thickness of the
material per unit area per unit temperature difference,
where the thermal conductivity of a material is a measure
of the ability of the material to conduct heat.

Note:- To measure thermal conductivity of a certain

material, we can follow the simple experimental setup
which is shown in Fig.(1.11), where a layer of material of
known thickness L and area A can be heated from one
side by an electric resistance heater of known output Q˙.
If the outer surfaces of the heater are well insulated, all
the heat generated by the resistance heater will be
transferred through the material whose conductivity is to
be determined. Then measuring the two surface
temperatures of the material when steady heat transfer is
reached and substituting them into Eq.(1.10) together
with other known quantities give the value of thermal
conductivity. Fig. (1.11) A simple experimental setup to
determine the thermal conductivity of a material

EXAMPLE 1.4-
A common way of measuring the thermal conductivity of a
material is to sandwich an electric thermo-foil heater between
two identical samples of the material, as shown in Fig.(1.12), a
circulating fluid such as tap water keeps the exposed ends of the
samples at constant temperature. The lateral surfaces of the
samples are well insulated to ensure that heat transfer through
the samples is one-dimensional. In a certain experiment,
cylindrical samples of diameter 5cm and length 10cm are used.
The two thermocouples in each sample are placed 3cm apart.
After initial transients, the electric heater is observed to draw
0.4A at 110V, and both differential thermometers read a
temperature difference of 15C°, Determine the thermal
conductivity of the sample.

SOLUTION
Assumptions 1 Steady operating conditions exist since the
temperature readings do not change with time. 2 Heat losses
through the lateral surfaces of the apparatus are negligible since
those surfaces are well insulated, and thus, the entire heat
generated by the heater is conducted through the samples. 3 Fig.(1.12) Schematic for Example-1.4
The apparatus possesses thermal symmetry.
Analysis The electrical power consumed by the resistance
heater and converted to heat is;
P=V×I=(110V)×(0.4A)=44W
The rate of heat flow through each sample is;
Q˙=P/2=22W
The area normal to the direction of heat flow is;
A=πD2/4=π(0.05m)2/4=0.00196 m2
Q˙=k A ΔT/L
k= Q˙ L/( A ΔT)=(22W)(0.03m)/[( 0.00196 m2)(15C°)
=22.4 W/m C°

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Chapter One Basics of Heat Transfer
Note1- The thermal conductivities of materials vary over a wide range, as shown in Fig.(1.13). The
thermal conductivities of gases such as air vary by a factor of 104 from those of pure metals such as
copper. Note that pure crystals and metals have the highest thermal conductivities, and gases and
insulating materials the lowest.
Note2- The thermal conductivity of an alloy of two metals is usually much lower than that of either
metal, as shown in Table (1.1). For example, the thermal conductivity of steel containing just 1% of
chrome is 62 W/m·C°, while the thermal conductivities of iron and chromium are 83 and 95 W/m·C°,
respectively.
Note3- The thermal conductivities of materials vary with temperature, where the variation of thermal
conductivity over certain temperature ranges is negligible for some materials, but significant for others,
as shown in Fig.(1.14).

Fig. (1.13) The range of thermal conductivity of Fig. (1.14) The variation of the thermal conductivity
various materials at room temperature of various solids, liquids, and gases with temperature

Table (1.1) The thermal conductivity of

an alloy is usually much lower than the
thermal conductivity of its components

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Chapter One Basics of Heat Transfer
1.2.1.b- Thermal Diffusivity Table (1.2) The thermal diffusivities of
The product ρCp, which is frequently encountered in heat transfer some materials at room temperature
analysis, is called the heat capacity of a material and represents the
heat storage capability of a material. Another material property that
appears in the transient heat conduction analysis is the thermal
diffusivity, which represents how fast heat diffuses through a
material and is defined as;
Heat Conducted k
= = (m 2 / s ) (1.12)
Heat Stored  Cp
The thermal diffusivities of some common materials at 20 C° are
given in Table (1.2).
Note- The thermal conductivity k represents how well a material
conducts heat, and the heat capacity ρCp represents how much
energy a material stores per unit volume. Therefore, the thermal
diffusivity of a material can be viewed as the ratio of the heat
conducted through the material to the heat stored per unit volume.
1.2.2- Convection
Convection is the mode of energy transfer between a solid
surface and the adjacent liquid or gas that is in motion, and it
involves the combined effects of conduction and fluid motion. The
faster the fluid motion, the greater the convection heat transfer. In
the absence of any bulk fluid motion, heat transfer between a solid
surface and the adjacent fluid is by pure conduction. Consider the
cooling of a hot block by blowing cool air over its top surface as
shown in Fig. (1.15). Energy is first transferred to the air layer
adjacent to the block by conduction, this energy is then carried away Fig. (1.15) Heat transfer from a hot
from the surface by convection. surface to air by convection

Note1- Convection is called forced convection if the fluid is forced

to flow over the surface by external means such as a fan, pump, or
the wind. In contrast, convection is called natural or (free)
convection if the fluid motion is caused by buoyancy forces that are
induced by density differences due to the variation of temperature
in the fluid Fig.(1.16).

Note2- The rate of convection heat transfer is observed to be

Fig. (1.16) Heat transfer from a hot
proportional to the temperature difference, and is conveniently surface to air by convection
expressed by Newton’s law of cooling as;
Table (1.3) Typical values of
Q Conv. = h As (Ts − T ) (W ) (1.13) convection heat transfer coefficient
where h is the convection heat transfer coefficient in W/m2·C° or
Btu/h·ft2·F°, As is the surface area through which convection heat
transfer takes place, Ts is the surface temperature, and T∞ is the
temperature of the fluid sufficiently far from the surface.

Note3- The convection heat transfer coefficient h is not a property

of the fluid. It is an experimentally determined parameter whose
value depends on all the variables influencing convection such as
the surface geometry, the nature of fluid motion, the properties of
the fluid, and the bulk fluid velocity. Typical values of h are given
in Table (1.3).

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Chapter One Basics of Heat Transfer

1 .5 -

Fig.(1.17).

Fig.(1.17) Schematic for

Example-1.5

1.2.3- Radiation
Radiation is the energy emitted by a body in the form of electromagnetic
waves as a result of the changes in the electronic configurations of the atoms
or molecules. Unlike conduction and convection, the transfer of energy by
radiation does not require the presence of a material medium to take place
as shown in Fig.(1.18). In fact, energy transfer by radiation is faster than it
by conduction or convection since it transfers at the speed of light. All
bodies at a temperature above absolute zero emit thermal radiation, where
the maximum rate of radiation that can be emitted from a surface at an
absolute temperature T1 in (K) to another surface at an absolute temperature
T2 is given by the Stefan–Boltzmann law as;
Qrad.max = 1  As1 (T1 − T2 )
4 4
(W ) (1.14)
where σ = 5.67 ×10-8 (W/m2
· K° 4)
is the Stefan–Boltzmann constant, while Fig.(1.18) A hot object in a
ε1 and As1 are the emissivity and surface area of the first radiative body, vacuum chamber loses heat by
respectively. radiation only

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Chapter One Basics of Heat Transfer

1.3- Units and Conversion Factors

Generally, all the engineering branches use the system of S.I. units mainly, but the need to deal with
English units will remain with us for many of next years. We therefore list some conversion factors from
English units to S.I. units as shown in Table(1.4) which provides multipliers for a selection of common
units. As an example of their use, we may convert a power of (1,000 Btu/hr) into (W) by using the given
multiplier;
1,000 Btu/hr × [0.29307 W/(Btu/hr)] ≈ 293.1 W

Note- The S.I. units may have prefixes placed front of them to indicate multiplication by various powers
of ten. For example, the prefix "k" denotes multiplication by 103. The complete set of S.I. prefixes is
given in Table(1.5).

Table (1.4) Selected Conversion Factors

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Chapter One Basics of Heat Transfer

Table (1.4) …continued

Table (1.5) S.I. Multiplying Factors

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Chapter One Basics of Heat Transfer
PROBLEMS constant pressure. Also, determine the cost of this
1.1- A cylindrical resistor element on a circuit heat if the unit cost of electricity in that area is
board dissipates 0.6 W of power. The resistor is 0.075$/kWh. Answers: 9038 kJ, 0.19 $
1.5cm long, and has a diameter of 0.4cm.
Assuming heat to be transferred uniformly from 1.5- Consider a 60-gallon water heater that is
all surfaces, determine (a) the amount of heat this initially filled with water at 45F°. Determine how
resistor dissipates during a 24-hour period, (b) the much energy needs to be transferred to the water
heat flux, and (c) the fraction of heat dissipated to raise its temperature to 140F°. Take the density
from the top and bottom surfaces. and specific heat of water to be 62 lbm/ft3 and 1.0
Answers: 51.84 kJ, 0.2809 W/m2, 11.8% Btu/lbm ·F°, respectively.
Answer: 47,250 Btu
3
1.6- A 1m rigid tank contains hydrogen at 250kPa
1.2- A 15-cm-diameter aluminum ball is to be
heated from 80C° to an average temperature of and 420K. The gas is now cooled until its
200C°. Taking the average density and specific temperature drops to 300K. Determine (a) the final
heat of aluminum in this temperature range to be pressure in the tank and (b) the amount of heat
ρ=2700kg/m3 and Cp=0.90kJ/kg·C°, respectively, transfer from the tank.
determine the amount of energy that needs to be Answers: 178.6 kPa, 180.0 kJ
transferred to the aluminum ball. 1.7- A 20kg mass of iron at 100C° is brought into
Answer: 515 kJ contact with 20kg of aluminum at 200C° in an
1.3- Consider an electrically heated house that has insulated enclosure. Determine the final
a floor space of 200m2 and an average height of equilibrium temperature of the combined system.
3m at 1000m elevation, where the standard Answer: 168C°
atmospheric pressure is 89.6kPa. The house is 1.8- An unknown mass of iron at 90C° is dropped
maintained at a temperature of 22C°, and the into an insulated tank that contains 80L of water at
infiltration losses are estimated to amount to 0.7 20C°. At the same time, a paddle wheel driven by
ACH. Assuming the pressure and the temperature a 200W motor is activated to stir the water.
in the house remain constant, determine the Thermal equilibrium is established after 25
amount of energy loss from the house due to minutes with a final temperature of 27C°. By
infiltration for a day during which the average neglecting the energy stored in the paddle wheel,
outdoor temperature is 5C°. Also, determine the determine the mass of the iron.
cost of this energy loss for that day if the unit cost Answer: 72.1kg
of electricity in that area is 0.082$/kWh. 1.9- A 90 lbm mass of copper at 160F° and a 50
Hint- (Infiltration of cold air into a warm house lbm mass of iron at 200F° are dropped into a tank
during winter through the cracks around doors, containing 180 lbm of water at 70F°. If 600 Btu of
windows, and other openings is a major source of heat is lost to the surroundings during the process,
energy loss since the cold air that enters needs to determine the final equilibrium temperature.
be heated to the room temperature. The infiltration Answer: 74.3F°
is often expressed in terms of ACH (air changes
1.10- The inner and outer surfaces of a 5m×6m
per hour). An ACH of 2 indicates that the entire air
brick wall of thickness 30cm and thermal
in the house is replaced twice every hour by the
conductivity 0.69 W/m ·C° are maintained at
cold air outside).
temperatures of 20C° and 5C°, respectively.
Answers: 53.8 kWh/day, 4.41$/day
Determine the rate of heat transfer through the
1.4- Consider a house with a floor space of 200m2
wall, in W. Answer: 1035 W
and an average height of 3m at sea level, where the
standard atmospheric pressure is 101.3kPa. 1.11- The inner and outer surfaces of a 0.5cm-
Initially the house is at a uniform temperature of thick 2m×2m window glass in winter are 10C° and
10C°. Now the electric heater is turned on, and the 3C°, respectively. If the thermal conductivity of
heater runs until the air temperature in the house the glass is 0.78 W/m ·C°, determine the amount
rises to an average value of 22C°. Determine how of heat loss, in kJ, through the glass over a period
much heat is absorbed by the air assuming some of 5 hours. What would your answer be if the glass
air escapes through the cracks as the heated air were 1cm-thick?
in the house expands at Answers: 78,624 kJ, 39,312 kJ

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Chapter One Basics of Heat Transfer
1.12- During an experiment, two 0.5-cm-thick Answers: 12.632kW, 0.038 kg/sec
samples 10cm×10cm in size are used. When
steady operation is reached, the heater is observed
to draw 35W of electric power, and the
temperature of each sample is observed to drop
from 82C° at the inner surface to 74C° at the outer
surface. Determine the thermal conductivity of the Prob.(1.16)
material at the average temperature.
Answer: 1.09W/m ∙C°
1.17- An ice chest whose outer dimensions are
30cm×40cm×40cm is made of 3cm-thick
Styrofoam (k = 0.033 W/m ·C°). Initially, the chest
is filled with 40kg of ice at 0C°, and the inner
surface temperature of the ice chest can be taken
Prob.(1.12) to be 0C° at all times. The heat of fusion of ice at
0C° is 333.7 kJ/kg, and the surrounding ambient
air is at 30C°. Disregarding any heat transfer from
the 40cm×40cm base of the ice chest, determine
1.13- Consider a person standing in a room how long it will take for the ice in the chest to melt
maintained at 20C° at all times. The inner surfaces completely if the outer surfaces of the ice chest are
of the walls, floors, and ceiling of the house are at 8C°.
observed to be at an average temperature of 12C° Answer: 32.7 days
in winter and 23C° in summer. Determine the rates
of radiation heat transfer between this person and
the surrounding surfaces in both summer and
winter if the exposed surface area, emissivity, and
the average outer surface temperature of the
person are 1.6m2, 0.95, and 32C°, respectively.
Answers: 84.2 W, 177.2 W Prob.(1.17)
1.14- Hot air at 80C° is blown over a 2m×4m flat
surface at 30C°. If the average convection heat
transfer coefficient is 55 W/m2 ·C°, determine the 1.18- Consider a sealed 20cm-high electronic box
rate of heat transfer from the air to the plate, in kW. whose base dimensions are 40cm×40cm placed in
Answer: 22 kW a vacuum chamber. The emissivity of the outer
1.15- A 50cm-long, 800W electric resistance surface of the box is 0.95. If the electronic
heating element with diameter 0.5cm and surface components in the box dissipate a total of 100 W
temperature 120C° is immersed in 60kg of water of power and the outer surface temperature of the
initially at 20C°. Determine how long it will take box is not to exceed 55C°, determine the
for this heater to raise the water temperature to temperature at which the surrounding surfaces
80C°. Also, determine the convection heat transfer must be kept if this box is to be cooled by radiation
coefficients at the beginning and at the end of the alone. Assume the heat transfer from the bottom
heating process if the specific heat of water is 4.18 surface of the box to the stand to be negligible.
kJ/kg∙C°. Answer: 23.3C°
Answers: 5.225hr, 1020W/m2∙C°, 2550W/m2∙C°
1.16- A hollow spherical iron container with outer
diameter 20cm and thickness 0.4cm is filled with
iced water at 0C°. If the outer surface temperature
is 5C°, determine the approximate rate of heat loss
from the sphere, in kW, and the rate at which ice
melts in the container. The heat of fusion of ice at Prob.(1.18)
0C° is 333.7 kJ/kg.

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