## 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**## 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

## 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).

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.

**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

## 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**## 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 ?

## 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**## Discussion

**NEW**

*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**## 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.

## 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.

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.

**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.

## 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**## 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

## 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.## 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.