Assignment 11 - Temperature and Heat
Crime Scene
At a crime scene, the forensic inestigator notes that 9.7-g lead bullet that was stopped in a door-frame apparently melted completely on impact. Assuming the bullet was fired at room temperature of 20°C, what does the investigator calculate the minimum muzzle velocity of the gun was?
360.3 m/s
360.3 m/s
Discussion
NEW
help?
I'm not even sure where to begin. I know the temperature
change since lead melts at 327 degC and the room temp is 20
degC, but what equation should I be using to find the speed?
NEW
Solution
This one is kind of a trick question to me cause my numbers
were not needed, my answer was the same as the one below.
Specific Heat of Lead = Cl = 130 J/kg*degC
Lf = Latent Heat of Fusion of Lead = 25 kj/kg * 1000 j
=25000 kj/kg
Tf = melting point of lead which is 327 deg C
Q gain = [Cl * m * (Tf-Ti)] + [m*Lf] = 1/2 mv^2
Mass is in all three terms so it gets canceled out and you
simply have:
QGain = Cl(Tf-Ti) + Lf = 1/2 v^2
Solve for V
(130 J/kg*degC)(327-20)+25,000 kkg = 1/2v^2
V=360.3 m/s
help?
I'm not even sure where to begin. I know the temperature
change since lead melts at 327 degC and the room temp is 20
degC, but what equation should I be using to find the speed?
NEW
Solution
This one is kind of a trick question to me cause my numbers
were not needed, my answer was the same as the one below.
Specific Heat of Lead = Cl = 130 J/kg*degC
Lf = Latent Heat of Fusion of Lead = 25 kj/kg * 1000 j
=25000 kj/kg
Tf = melting point of lead which is 327 deg C
Q gain = [Cl * m * (Tf-Ti)] + [m*Lf] = 1/2 mv^2
Mass is in all three terms so it gets canceled out and you
simply have:
QGain = Cl(Tf-Ti) + Lf = 1/2 v^2
Solve for V
(130 J/kg*degC)(327-20)+25,000 kkg = 1/2v^2
V=360.3 m/s
Boiling Water in an Iron Boiler
An iron boiler of mass 233 kg contains 800 kg of water at 19.4°C. A heater supplies energy at the rate of 43400 kJ/hr. How long does it take for the water to reach the boiling point? Specific heat of iron = 450 J/(kg·K).
6.41 hr
How long does it take for the water to all have changed to steam? Latent heat of vaporization of water is Lv = 2260 kJ/kg.
4.81×101 hr
6.41 hr
How long does it take for the water to all have changed to steam? Latent heat of vaporization of water is Lv = 2260 kJ/kg.
4.81×101 hr
Discussion
NEW
Solution
First, lets get some values out of the way:
mass of water (mw)= 837kg mass of iron (mi)= 276
sp heat of water (cw)= 1.0 kcal/ kg*C sp heat of iron (ci)=
.11cal/kg*C
heat of vaporization of water (Lw)= 539 kcal/kg
W= 55800kJ/hr (convert to joules/ sec) * 1000J/kJ *
1hr/3600sec = 15500 J/s
BP of water = 100C
Temp of water (given in lon capa)= 21.4
so now the first part:
since there is NO PHASE CHANGE (meaning the water stays a
liquid) you use: mc(delta T)
(mw + mc)(delta T) = Q
[(276kg * .11cal/kg*C) + (837kg * 1.0 kcal/kg*C)] (100C -
21.4)= 68174.5 kcal
Now we must use this info in the Work equation: W= J/s and
since we need seconds, s= J/W
kcal can be converted to joules by multiplying by 4186J, so
J= 68174.5 * 4186 = 2.8e8J
s= 2.8e8J/ 15500W = 18412 seconds or 5.11 hr lon capa will
accept both
--------------------------------second part:
since there is a phase change in which the water evaporates,
we now have to think about the total Q, which is calculated
by examining each "step"
liquid water @ 21.4C --> liquid water @ 100C
Q1= mc(deltaT) (we already have this from problem 1 (68174.5
kcal))
liquid water @ 100C --> gas water @ 100C
Q2= m of water * latent heat of water
*****don't worry about the iron in the second step because
iron does not change its "phase"
Q2= 837kg (539 kcal/ kg) = 451143 kcal
Q1 + Q2 = 451143 kcal + 68174.5 kcal = 519317.5 kcal *
4186J/kcal = 2.2e9 Joules
s= J/W = 2173863055 J/ 15500 W = 1.40e5 seconds or 39 hr
Solution
First, lets get some values out of the way:
mass of water (mw)= 837kg mass of iron (mi)= 276
sp heat of water (cw)= 1.0 kcal/ kg*C sp heat of iron (ci)=
.11cal/kg*C
heat of vaporization of water (Lw)= 539 kcal/kg
W= 55800kJ/hr (convert to joules/ sec) * 1000J/kJ *
1hr/3600sec = 15500 J/s
BP of water = 100C
Temp of water (given in lon capa)= 21.4
so now the first part:
since there is NO PHASE CHANGE (meaning the water stays a
liquid) you use: mc(delta T)
(mw + mc)(delta T) = Q
[(276kg * .11cal/kg*C) + (837kg * 1.0 kcal/kg*C)] (100C -
21.4)= 68174.5 kcal
Now we must use this info in the Work equation: W= J/s and
since we need seconds, s= J/W
kcal can be converted to joules by multiplying by 4186J, so
J= 68174.5 * 4186 = 2.8e8J
s= 2.8e8J/ 15500W = 18412 seconds or 5.11 hr lon capa will
accept both
--------------------------------second part:
since there is a phase change in which the water evaporates,
we now have to think about the total Q, which is calculated
by examining each "step"
liquid water @ 21.4C --> liquid water @ 100C
Q1= mc(deltaT) (we already have this from problem 1 (68174.5
kcal))
liquid water @ 100C --> gas water @ 100C
Q2= m of water * latent heat of water
*****don't worry about the iron in the second step because
iron does not change its "phase"
Q2= 837kg (539 kcal/ kg) = 451143 kcal
Q1 + Q2 = 451143 kcal + 68174.5 kcal = 519317.5 kcal *
4186J/kcal = 2.2e9 Joules
s= J/W = 2173863055 J/ 15500 W = 1.40e5 seconds or 39 hr
Specific Heat of A Metal
A metal container, which has a mass of 7.0 kg contains 17.3 kg of water. A 2.9-kg piece of the same metal, initially at a temperature of 170.0°C, is dropped into the water. The container and the water initially have a temperature of 17.0°C and the final temperature of the entire system is 19.0°C. Calculate the specific heat of the metal.
342 J/kg/K
342 J/kg/K
Discussion
NEW
units
What are they units for this question/ how do you enter
them in. I tried J / Kg*C and kJ/kG*C and neither work
NEW
Re: units Anonymous 2
should be J/(kg*degC)
NEW
Anonymous 5
The units for specific heat should be entered as J/kg/K.
NEW
Anonymous 4
what equation do you use for this?
NEW
kg of water=17.4kg
heat constant=4200
sample metal=1.9kg
metal container=9.2kg
initial temp=14.9C
final temp=16.9C
initial temp dropped in the water=160.0C
C=4200 * 17.4 (16.9-14.9)/(1.9)(160.0-16.9)-(9.2*16.9-14.9)
units J/kg/K
NEW
To
You are setting your equation up completely wrong
Qgain = Qlost. we just did this in lab this week.
I assume by heat constant you mean the specific heat of
water? Which is 4180 J/kgdegC, not 4200
Here, let X = the unknown specific heat of your sample,
also remember that the container and the sample are made of
the same material, and thus have the same specific heat.
Mconatiner*X*(Tf-Ti) + Mwater*4180*(Tf-Ti) = Msample*X*(Ti
of sample - Tf)
ran your numbers fast on my calc, assuming I didnt make a
simple mistake you should come up with a value around 518.3
J/(kg*degC)
NEW
Re:
how do you put this into units of J/kg/K ?
units
What are they units for this question/ how do you enter
them in. I tried J / Kg*C and kJ/kG*C and neither work
NEW
Re: units Anonymous 2
should be J/(kg*degC)
NEW
Anonymous 5
The units for specific heat should be entered as J/kg/K.
NEW
Anonymous 4
what equation do you use for this?
NEW
kg of water=17.4kg
heat constant=4200
sample metal=1.9kg
metal container=9.2kg
initial temp=14.9C
final temp=16.9C
initial temp dropped in the water=160.0C
C=4200 * 17.4 (16.9-14.9)/(1.9)(160.0-16.9)-(9.2*16.9-14.9)
units J/kg/K
NEW
To
You are setting your equation up completely wrong
Qgain = Qlost. we just did this in lab this week.
I assume by heat constant you mean the specific heat of
water? Which is 4180 J/kgdegC, not 4200
Here, let X = the unknown specific heat of your sample,
also remember that the container and the sample are made of
the same material, and thus have the same specific heat.
Mconatiner*X*(Tf-Ti) + Mwater*4180*(Tf-Ti) = Msample*X*(Ti
of sample - Tf)
ran your numbers fast on my calc, assuming I didnt make a
simple mistake you should come up with a value around 518.3
J/(kg*degC)
NEW
Re:
how do you put this into units of J/kg/K ?
Melting Ice
0.135 kg of water at 87.0°C is poured into an insulated cup containing 0.527 kg of ice initially at 0°C. How many kg of liquid will there be when the system reaches thermal equilibrium?
0.283 kg
0.283 kg
Discussion
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Anonymous 1
how do I start this problem?
NEW
Anonymous 2
does anyone know how to start this problem?
NEW
Anonymous 3
any help on this??
NEW
Figured it out
Assume the final Temperature is 0 degrees C, and it becomes
quite simple. 334000*m = mass water * 4186 * Ti , then
divide by 334000
my numbers were .160 kg water at 85 C
and 0.563 kg ice at 0
What is interesting, is that if the water at 85 degrees
does not have enough heat to change all the ice to liquid,
then the temperature will still be zero, from what I can
ascertain at least.
so, 0.160*4186*85 = 56929.6 J This is my total amount
of energy available to melt ice, or raise the temperature.
it would take, however, 334000*0.563 = 188942 J to melt all
the ice to liquid, clearly this cannot happen since my Q
from water (56929) is less than 188942.
remember Qlost = Qgain?
My Q lost is the 56929.6J, and thus Q gain is the same. so
the question becomes how much of a 0.563Kg block of ice
will melt? simple, 56929.6 J / 334000 J = mass of ice
that will melt, here, = 0.1704 kg, add that back to the
initial mass of water, 0.1704 + 0.160 = 0.330kg total
liquid after thermal equilibrium
NEW
adding to the above
note that your inital weight of ice is irrelevant, you are
simply finding how much ice your water is capable of
melting.
mass water * 4186 * initial T water / 334000 = mass of
ice that will become liquid
add that mass to your inital mass of water and you have
your answer
Anonymous 1
how do I start this problem?
NEW
Anonymous 2
does anyone know how to start this problem?
NEW
Anonymous 3
any help on this??
NEW
Figured it out
Assume the final Temperature is 0 degrees C, and it becomes
quite simple. 334000*m = mass water * 4186 * Ti , then
divide by 334000
my numbers were .160 kg water at 85 C
and 0.563 kg ice at 0
What is interesting, is that if the water at 85 degrees
does not have enough heat to change all the ice to liquid,
then the temperature will still be zero, from what I can
ascertain at least.
so, 0.160*4186*85 = 56929.6 J This is my total amount
of energy available to melt ice, or raise the temperature.
it would take, however, 334000*0.563 = 188942 J to melt all
the ice to liquid, clearly this cannot happen since my Q
from water (56929) is less than 188942.
remember Qlost = Qgain?
My Q lost is the 56929.6J, and thus Q gain is the same. so
the question becomes how much of a 0.563Kg block of ice
will melt? simple, 56929.6 J / 334000 J = mass of ice
that will melt, here, = 0.1704 kg, add that back to the
initial mass of water, 0.1704 + 0.160 = 0.330kg total
liquid after thermal equilibrium
NEW
adding to the above
note that your inital weight of ice is irrelevant, you are
simply finding how much ice your water is capable of
melting.
mass water * 4186 * initial T water / 334000 = mass of
ice that will become liquid
add that mass to your inital mass of water and you have
your answer
Steam to Ice
How many J of energy must be removed when 122.0 g of steam, at a temperature of 166.0°C, is cooled and frozen into 122.0 g of ice at 0°C? Take the specific heat of steam to be 2.1 kJ/(kg·K).
3.84×105 J
3.84×105 J
Discussion
NEW
what is wrong
I am trying to do this problem and taking into account 4
things:
changing from steam to water (m*csteam*deltaT)where cstaem
= 2.1
Q condensation (-m*Lvaporization)where Lvap = 2260
Q of cooling water (m*cwater*deltaT)where cwater = 4.186
and Q of freezing all the water (-m*Lfusion) where Lf = 333
what am I doing wrong?
Thanks!
NEW
Natalia Tobon
specific heat of water is 4.186 kJ/kgC
heat of fusion of ice is 334 kJ/kg
Assuming that not all of the ice melts, then the water cools
from 92 to 0, releasing this energy:
E = 4.186 kJ/kgC x 0.125 kg x 92C = 48.1 kJ
How much ice can this melt?
48.1 kJ = 334 kJ/kg x m
m = 0.144 kg
So 0.144 kg of ice is melted, and total amount of water is
0.144 + 0.125 = 0.269 kg
NEW
Sorry ignore last post that was for the previous problem
this one is solved by steps since there are 2 phase changes.
First find the heat required to make the steam into water.
The subtract 540 J/g because that is the heat released
during the phase change.
q = mcΔT
q = (148.0 g)(2.1 J/g C)(100-200 C)
q = -31080 J
540 J/g x 148 g = 79920 J
q = -31080 - 79920 = -111000 J
Now find the heat release when water turns to ice. Again
subtract 80 J/g released when going from liquid to ice.
q = mcΔT
q = (148.0 g)(4.184 J/g C)(0-100 C)
q = -61923.2 J
80 J/g x 148 g = 11840 J
q = -61923.2 - 11840 = -73763.2 J
Now add the two values of q together to find the total.
q = -111000 + -73763.2 = -184763.2 J
NEW
Anonymous 6
can someone help me out with this problem? I tried the
previous post and its not working for me, I only have one
try left please help!
NEW
help?
I've tried the previous help post and am still getting the
wrong answer and still rather confused :/ help?
NEW
i keep trying it but i cant seem to get the right answer
does anyone else know how to do this problem?
NEW
Anonymous 9
merrrr. can soemone pleaes post how to do this haha.
thank youuuuu =)
NEW
Answer isnt negative!!!!
its so easy
but the answer is positive!
2.1 * mass * delta t (given T to 100 degress)
+
2200 * mass
+
4.184 * mass * 100 degress (delta t from steam to 0, which
is 100 degrees)
+
334 * mass
total in kJ but not negative, its positive!
NEW
and as for the post the girl above made, she listed the
latent heats in calories, not in joules, those answers
would be incorrect.
you would have to mulitply 540(cal, not j) * mass * 4.186
(cal to joule conversion)
effectively, it makes more sense to simply multiply the
latent heat of fusion or vaporization that we are given to
start with, rather than needlessly converting between
calories and joules
2260 * mass is latent heat of vap
334 * mass is latent heat of fusion, these answers will
be in kj/kj so be sure to change your figures according
to your units.. i.e.; convert grams to kg or down
convert 2260 kj per kg to 2.260 kj per gram
NEW
adding to above
though, i do commend miss natalia on her clear
understanding of the material, she just forgot to list the
conversion, an honest mistake.
what is wrong
I am trying to do this problem and taking into account 4
things:
changing from steam to water (m*csteam*deltaT)where cstaem
= 2.1
Q condensation (-m*Lvaporization)where Lvap = 2260
Q of cooling water (m*cwater*deltaT)where cwater = 4.186
and Q of freezing all the water (-m*Lfusion) where Lf = 333
what am I doing wrong?
Thanks!
NEW
Natalia Tobon
specific heat of water is 4.186 kJ/kgC
heat of fusion of ice is 334 kJ/kg
Assuming that not all of the ice melts, then the water cools
from 92 to 0, releasing this energy:
E = 4.186 kJ/kgC x 0.125 kg x 92C = 48.1 kJ
How much ice can this melt?
48.1 kJ = 334 kJ/kg x m
m = 0.144 kg
So 0.144 kg of ice is melted, and total amount of water is
0.144 + 0.125 = 0.269 kg
NEW
Sorry ignore last post that was for the previous problem
this one is solved by steps since there are 2 phase changes.
First find the heat required to make the steam into water.
The subtract 540 J/g because that is the heat released
during the phase change.
q = mcΔT
q = (148.0 g)(2.1 J/g C)(100-200 C)
q = -31080 J
540 J/g x 148 g = 79920 J
q = -31080 - 79920 = -111000 J
Now find the heat release when water turns to ice. Again
subtract 80 J/g released when going from liquid to ice.
q = mcΔT
q = (148.0 g)(4.184 J/g C)(0-100 C)
q = -61923.2 J
80 J/g x 148 g = 11840 J
q = -61923.2 - 11840 = -73763.2 J
Now add the two values of q together to find the total.
q = -111000 + -73763.2 = -184763.2 J
NEW
Anonymous 6
can someone help me out with this problem? I tried the
previous post and its not working for me, I only have one
try left please help!
NEW
help?
I've tried the previous help post and am still getting the
wrong answer and still rather confused :/ help?
NEW
i keep trying it but i cant seem to get the right answer
does anyone else know how to do this problem?
NEW
Anonymous 9
merrrr. can soemone pleaes post how to do this haha.
thank youuuuu =)
NEW
Answer isnt negative!!!!
its so easy
but the answer is positive!
2.1 * mass * delta t (given T to 100 degress)
+
2200 * mass
+
4.184 * mass * 100 degress (delta t from steam to 0, which
is 100 degrees)
+
334 * mass
total in kJ but not negative, its positive!
NEW
and as for the post the girl above made, she listed the
latent heats in calories, not in joules, those answers
would be incorrect.
you would have to mulitply 540(cal, not j) * mass * 4.186
(cal to joule conversion)
effectively, it makes more sense to simply multiply the
latent heat of fusion or vaporization that we are given to
start with, rather than needlessly converting between
calories and joules
2260 * mass is latent heat of vap
334 * mass is latent heat of fusion, these answers will
be in kj/kj so be sure to change your figures according
to your units.. i.e.; convert grams to kg or down
convert 2260 kj per kg to 2.260 kj per gram
NEW
adding to above
though, i do commend miss natalia on her clear
understanding of the material, she just forgot to list the
conversion, an honest mistake.
Thermal Conductivity of Down Feather
A mountain climber wears down clothing 3.30 cm thick with total surface area 1.61 m2. The temperature at the surface of the clothing is -18.1°C and at the skin is 32.4°C. Determine the rate of heat flow by conduction through the clothing assuming it is dry and that the thermal conductivity, k, is that of down.
6.16×101 W
Determine the rate of heat flow by conduction through the clothing assuming the clothing is wet, so that k is that of water and the jacket has matted down to 0.524 cm thickness.
8.69×103 W
6.16×101 W
Determine the rate of heat flow by conduction through the clothing assuming the clothing is wet, so that k is that of water and the jacket has matted down to 0.524 cm thickness.
8.69×103 W
Discussion
NEW
Re:
k = 0.025 J/smk
NEW
Re:
and k = 0.025 for water
NEW
part 1 Anonymous 3
I am doing the equation: KA(deltaT/deltaX), where K=.025,
A=the area, delta T=change in temps, delta X=distance of
down to skin. I get a number around 58, but not sure what
the units should be or if I need to convert to something
else.
NEW
part2
wouldn't you use the same formula and just change the thickness?
NEW
Re: part 1 Anonymous 5
Make sure that your thickness is in the same units as your A.
My answer was in the 50's also. The unit I used was J/s
NEW
Re: Re: part 1
This is assuming you're talking about the first problem.
NEW
Anonymous 6
Can someone please help me with part 2?
NEW
K water Anonymous 5
is 0.561 J/smk by the way, not what was posted.
The k of down is correct, though
NEW
Re: part2 Anonymous 7
For part 2 use the same formula but you change the k to
0.561 (the k of water) and the thickness to what you were
given...this worked for me.
NEW
Re: part 1
can somebody help me as well im having the same problem...i
converted my thickness from cm to m and im not getting the
correct answer: 3.37cm=0.0337m
Q/t=kA(delta T/L)
=(0.025 W/K*m)(1.72 m^2)((34.8-20.6)/0.0337 m)=18.12 W
NEW
Anonymous 9
http://physics.hivepc.com/thermo.html
Question #11
NEW
Re: Re: part 1
ok i got it for the temperature you are not supposes to
subtract you add, for both questions every thing else you
are doing is correct. of course remember for 2nd question
the k is different.
Re:
k = 0.025 J/smk
NEW
Re:
and k = 0.025 for water
NEW
part 1 Anonymous 3
I am doing the equation: KA(deltaT/deltaX), where K=.025,
A=the area, delta T=change in temps, delta X=distance of
down to skin. I get a number around 58, but not sure what
the units should be or if I need to convert to something
else.
NEW
part2
wouldn't you use the same formula and just change the thickness?
NEW
Re: part 1 Anonymous 5
Make sure that your thickness is in the same units as your A.
My answer was in the 50's also. The unit I used was J/s
NEW
Re: Re: part 1
This is assuming you're talking about the first problem.
NEW
Anonymous 6
Can someone please help me with part 2?
NEW
K water Anonymous 5
is 0.561 J/smk by the way, not what was posted.
The k of down is correct, though
NEW
Re: part2 Anonymous 7
For part 2 use the same formula but you change the k to
0.561 (the k of water) and the thickness to what you were
given...this worked for me.
NEW
Re: part 1
can somebody help me as well im having the same problem...i
converted my thickness from cm to m and im not getting the
correct answer: 3.37cm=0.0337m
Q/t=kA(delta T/L)
=(0.025 W/K*m)(1.72 m^2)((34.8-20.6)/0.0337 m)=18.12 W
NEW
Anonymous 9
http://physics.hivepc.com/thermo.html
Question #11
NEW
Re: Re: part 1
ok i got it for the temperature you are not supposes to
subtract you add, for both questions every thing else you
are doing is correct. of course remember for 2nd question
the k is different.
Planet - Average Surface Temperature
The planet, Saturn, receives about 4.686 W/m2 from the Sun, averaged over the whole surface, and radiates an equal amount back into space (that is Saturn is in equilibrium). Assuming Saturn is a perfect emitter (e = 1.0) estimate its average surface temperature.
95.3 K
95.3 K
Discussion
NEW
Solution!!!
The planet, Venus, receives about 861 W/m2 from the Sun,
averaged over the whole surface, and radiates an equal
amount back into space (that is Venus is in equilibrium).
Assuming Venus is a perfect emitter (e = 1.0) estimate its
average surface temperature.
The radiation is due to the thermodynamic black body so this
equation deals with the stephan- boltzmann constant on pg 399.
So, the equation is the radiation that Venus recieves from
the sun which is,
861 w/m^2 = ( 5.67e-8 w/m^2 *K^4)
rearrange the equation so,
the radation recieved from the sun, divided by the stefan-
boltzmann equation all raised to the 1/4th power.
T= (861/5.67e-8))^(1/4)
T= (1.518518519 e10)^(1/4)
answer: 351.0 K
NEW
Re: Solution!!! Anonymous 2
I've done this calculation several times, and it's not
working for me. I even used your numbers using the formula
you posted, and your formula and numbers doesn't match the
solution you posted.
Could you clarify this please?
NEW
Answer
You're trying to find a temperature value (measured in K).
Step 1:
Your value in Watts/Stefan-B. constant: 5.67e-8
--> (15.5 W/m^2)/(5.67e-8)= 2.73e8
Step 2:
Take answer to the 1/4
--> (2.73e8)^(1/4)= 128.6K
Solution!!!
The planet, Venus, receives about 861 W/m2 from the Sun,
averaged over the whole surface, and radiates an equal
amount back into space (that is Venus is in equilibrium).
Assuming Venus is a perfect emitter (e = 1.0) estimate its
average surface temperature.
The radiation is due to the thermodynamic black body so this
equation deals with the stephan- boltzmann constant on pg 399.
So, the equation is the radiation that Venus recieves from
the sun which is,
861 w/m^2 = ( 5.67e-8 w/m^2 *K^4)
rearrange the equation so,
the radation recieved from the sun, divided by the stefan-
boltzmann equation all raised to the 1/4th power.
T= (861/5.67e-8))^(1/4)
T= (1.518518519 e10)^(1/4)
answer: 351.0 K
NEW
Re: Solution!!! Anonymous 2
I've done this calculation several times, and it's not
working for me. I even used your numbers using the formula
you posted, and your formula and numbers doesn't match the
solution you posted.
Could you clarify this please?
NEW
Answer
You're trying to find a temperature value (measured in K).
Step 1:
Your value in Watts/Stefan-B. constant: 5.67e-8
--> (15.5 W/m^2)/(5.67e-8)= 2.73e8
Step 2:
Take answer to the 1/4
--> (2.73e8)^(1/4)= 128.6K
Insulating Properties
A goose-down coat keeps you warm. Why? For each of the following statements determine whether it is true or false.
True: The heat that keeps you warm comes from your body.
True: A wet goose down is a less good insulator because water has a moderate thermal conductivity.
False: The heat that keeps you warm comes from the outside air.
False: The heat that keeps you warm comes from goose down.
True: The goose down is a very good insulator.
True: The heat that keeps you warm comes from your body.
True: A wet goose down is a less good insulator because water has a moderate thermal conductivity.
False: The heat that keeps you warm comes from the outside air.
False: The heat that keeps you warm comes from goose down.
True: The goose down is a very good insulator.
Discussion
NEW
Anonymous 1
Any help? Im on my last try and this is what I have...any
help would be appreciated!
A goose-down coat keeps you warm. Why? For each of the
following statements determine whether it is true or false.
False The heat that keeps you warm comes from the outside
air.
False A wet goose down is a less good insulator because
the water has a large specific heat.
True The goose down is a very good insulator.
False The heat that keeps you warm comes from goose down.
False A wet goose down is a less good insulator because
water has a moderate thermal conductivity.
NEW
Answers
True: A wet goose down is a less good insulator because
water has a moderate thermal conductivity.
False: The heat that keeps you warm comes from goose down.
True: The heat that keeps you warm comes from your body.
True: The goose down is a very good insulator.
False: A wet goose down is a less good insulator because the
water has a large specific heat.
NEW
answers
True: A wet goose down is a less good insulator because
water has a moderate thermal conductivity.
True: The goose down is a very good insulator.
Anonymous 1
Any help? Im on my last try and this is what I have...any
help would be appreciated!
A goose-down coat keeps you warm. Why? For each of the
following statements determine whether it is true or false.
False The heat that keeps you warm comes from the outside
air.
False A wet goose down is a less good insulator because
the water has a large specific heat.
True The goose down is a very good insulator.
False The heat that keeps you warm comes from goose down.
False A wet goose down is a less good insulator because
water has a moderate thermal conductivity.
NEW
Answers
True: A wet goose down is a less good insulator because
water has a moderate thermal conductivity.
False: The heat that keeps you warm comes from goose down.
True: The heat that keeps you warm comes from your body.
True: The goose down is a very good insulator.
False: A wet goose down is a less good insulator because the
water has a large specific heat.
NEW
answers
True: A wet goose down is a less good insulator because
water has a moderate thermal conductivity.
True: The goose down is a very good insulator.