# 1.6 Mechanisms of heat transfer  (Page 16/27)

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You leave a pastry in the refrigerator on a plate and ask your roommate to take it out before you get home so you can eat it at room temperature, the way you like it. Instead, your roommate plays video games for hours. When you return, you notice that the pastry is still cold, but the game console has become hot. Annoyed, and knowing that the pastry will not be good if it is microwaved, you warm up the pastry by unplugging the console and putting it in a clean trash bag (which acts as a perfect calorimeter) with the pastry on the plate. After a while, you find that the equilibrium temperature is a nice, warm $38.3\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ . You know that the game console has a mass of 2.1 kg. Approximate it as having a uniform initial temperature of $45\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ . The pastry has a mass of 0.16 kg and a specific heat of $3.0\phantom{\rule{0.2em}{0ex}}\text{k}\phantom{\rule{0.2em}{0ex}}\text{J/}\left(\text{kg}·\text{ºC}\right),$ and is at a uniform initial temperature of $4.0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ . The plate is at the same temperature and has a mass of 0.24 kg and a specific heat of $0.90\phantom{\rule{0.2em}{0ex}}\text{J/}\left(\text{kg}·\text{ºC}\right)$ . What is the specific heat of the console?

$1.7\phantom{\rule{0.2em}{0ex}}\text{kJ/}\left(\text{kg}·\text{ºC}\right)$

Two solid spheres, A and B , made of the same material, are at temperatures of $0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ and $100\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ , respectively. The spheres are placed in thermal contact in an ideal calorimeter, and they reach an equilibrium temperature of $20\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ . Which is the bigger sphere? What is the ratio of their diameters?

In some countries, liquid nitrogen is used on dairy trucks instead of mechanical refrigerators. A 3.00-hour delivery trip requires 200 L of liquid nitrogen, which has a density of $808\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}.$ (a) Calculate the heat transfer necessary to evaporate this amount of liquid nitrogen and raise its temperature to $3.00\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ . (Use ${c}_{\text{P}}$ and assume it is constant over the temperature range.) This value is the amount of cooling the liquid nitrogen supplies. (b) What is this heat transfer rate in kilowatt-hours? (c) Compare the amount of cooling obtained from melting an identical mass of $0\text{-}\text{°}\text{C}$ ice with that from evaporating the liquid nitrogen.

a. $1.57\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{kcal}$ ; b. $18.3\phantom{\rule{0.2em}{0ex}}\text{kW}·\text{h}$ ; c. $1.29\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{kcal}$

Some gun fanciers make their own bullets, which involves melting lead and casting it into lead slugs. How much heat transfer is needed to raise the temperature and melt 0.500 kg of lead, starting from $25.0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ ?

A 0.800-kg iron cylinder at a temperature of $1.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ is dropped into an insulated chest of 1.00 kg of ice at its melting point. What is the final temperature, and how much ice has melted?

$6.3\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ . All of the ice melted.

Repeat the preceding problem with 2.00 kg of ice instead of 1.00 kg.

Repeat the preceding problem with 0.500 kg of ice, assuming that the ice is initially in a copper container of mass 1.50 kg in equilibrium with the ice.

$63.9\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ , all the ice melted

A 30.0-g ice cube at its melting point is dropped into an aluminum calorimeter of mass 100.0 g in equilibrium at $24.0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ with 300.0 g of an unknown liquid. The final temperature is $4.0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ . What is the heat capacity of the liquid?

(a) Calculate the rate of heat conduction through a double-paned window that has a $1.50{\text{-m}}^{2}$ area and is made of two panes of 0.800-cm-thick glass separated by a 1.00-cm air gap. The inside surface temperature is $15.0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C},$ while that on the outside is $-10.0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}.$ ( Hint: There are identical temperature drops across the two glass panes. First find these and then the temperature drop across the air gap. This problem ignores the increased heat transfer in the air gap due to convection.) (b) Calculate the rate of heat conduction through a 1.60-cm-thick window of the same area and with the same temperatures. Compare your answer with that for part (a).

a. 83 W; b. $1.97\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{W}$ ; The single-pane window has a rate of heat conduction equal to 1969/83, or 24 times that of a double-pane window.

a body is projected vertically upward of 30kmp/h how long will it take to reach a point 0.5km bellow e point of projection
i have to say. who cares. lol. why know that t all
Jeff
is this just a chat app about the openstax book?
kya ye b.sc ka hai agar haa to konsa part
what is charge quantization
it means that the total charge of a body will always be the integral multiples of basic unit charge ( e ) q = ne n : no of electrons or protons e : basic unit charge 1e = 1.602×10^-19
Riya
is the time quantized ? how ?
Mehmet
What do you meanby the statement,"Is the time quantized"
Mayowa
Can you give an explanation.
Mayowa
there are some comment on the time -quantized..
Mehmet
time is integer of the planck time, discrete..
Mehmet
planck time is travel in planck lenght of light..
Mehmet
it's says that charges does not occur in continuous form rather they are integral multiple of the elementary charge of an electron.
Tamoghna
it is just like bohr's theory. Which was angular momentum of electron is intral multiple of h/2π
determine absolute zero
The properties of a system during a reversible constant pressure non-flow process at P= 1.6bar, changes from constant volume of 0.3m³/kg at 20°C to a volume of 0.55m³/kg at 260°C. its constant pressure process is 3.205KJ/kg°C Determine: 1. Heat added, Work done, Change in Internal Energy and Change in Enthalpy
U can easily calculate work done by 2.303log(v2/v1)
Abhishek
Amount of heat added through q=ncv^delta t
Abhishek
Change in internal energy through q=Q-w
Abhishek
please how do dey get 5/9 in the conversion of Celsius and Fahrenheit
what is copper loss
this is the energy dissipated(usually in the form of heat energy) in conductors such as wires and coils due to the flow of current against the resistance of the material used in winding the coil.
Henry
it is the work done in moving a charge to a point from infinity against electric field
what is the weight of the earth in space
As w=mg where m is mass and g is gravitational force... Now if we consider the earth is in gravitational pull of sun we have to use the value of "g" of sun, so we can find the weight of eaeth in sun with reference to sun...
Prince
g is not gravitacional forcé, is acceleration of gravity of earth and is assumed constante. the "sun g" can not be constant and you should use Newton gravity forcé. by the way its not the "weight" the physical quantity that matters, is the mass
Jorge
Yeah got it... Earth and moon have specific value of g... But in case of sun ☀ it is just a huge sphere of gas...
Prince
Thats why it can't have a constant value of g ....
Prince
not true. you must know Newton gravity Law . even a cloud of gas it has mass thats al matters. and the distsnce from the center of mass of the cloud and the center of the mass of the earth
Jorge
please why is the first law of thermodynamics greater than the second
every law is important, but first law is conservation of energy, this state is the basic in physics, in this case first law is more important than other laws..
Mehmet
First Law describes o energy is changed from one form to another but not destroyed, but that second Law talk about entropy of a system increasing gradually
Mayowa
first law describes not destroyer energy to changed the form, but second law describes the fluid drection that is entropy. in this case first law is more basic accorging to me...
Mehmet
define electric image.obtain expression for electric intensity at any point on earthed conducting infinite plane due to a point charge Q placed at a distance D from it.
explain the lack of symmetry in the field of the parallel capacitor
pls. explain the lack of symmetry in the field of the parallel capacitor
Phoebe
does your app come with video lessons?
What is vector
Vector is a quantity having a direction as well as magnitude
Damilare