Download Find the resulting acceleration from a 300 N force that acts on an

** SOLUTIONS Jaquin’s Sections **
SCI111 Homework Assignment #5
(20 Points Total)
Chapter 4
1. The Fahrenheit temperature reading is 98 degrees on a hot summer day. What is this
reading on the Kelvin scale?
This is a temperature scale conversion problem. Use the conversion formulae from the
Summary of Equations. You must first convert the degrees F into Degrees C, then
convert to Kelvin.
5
TF  32  5 98  32  5 66  36.7
9
9
9
TK  TC  273  36.7  2273  309.7
TC 
98 degrees F corresponds to 36.7 degrees C and 309.7 K.
3. A 1,000 kg car with a speed of 90 km/hr brakes to a stop. How many cal of heat are
generated by the brakes as a result?
This is a conservation of energy problem where kinetic energy is converted into heat. For
the car to come to stop all the kinetic energy must be bled away into some other form of energy,
in this case heat energy.
KE  Heat , Q
2
1 2 1
 km  1
 m
5 kg  m
Q  KE  mv  1,000kg 90
 3.125  10 5 J
  1,000kg 25   3.125  10
2
2
2
s
 hr  2
 s
2
The brakes must generate 312,000 Joules of energy to stop the car. This is equivalent to74,701
calories of heat.
6. A 50.0 g silver spoon at 20 degrees C is placed in a cup of coffee at 90.0 degrees C. How
much heat does the spoon absorb from the coffee to reach a temperature of 89.0 degrees C?
This problem is tricky because it gives you information that is not needed to solve the problem.
Since we are asked only how much heat the spoon absorbs to warm up 69 degrees C, the
information about the coffee is completely irrelevant to the problem. This is a simple Q=mcT
problem.
Q  mcT  50 g  0.056
cal
 69  C  193.2cal

g C
** SOLUTIONS Jaquin’s Sections **
SCI111 Homework Assignment #5
(20 Points Total)
The spoon absorbs 193.2 calories of heat.
8. How many minutes would be required for a 300 Watt immersion heater to heat 250.0 g
of water from 20 degrees C to 100 degrees C?
There are two ways to do this problem. Both are instructive. First, simply calculate the
heat necessary to raise the aquarium water temperature to the desired amount using Q=mcT, and
then determine the time required for the heated to deliver that much energy using the definition
of power P=Energy/time.
cal
 80  C  20,000cal  83,680 J

g C
Energy
Energy 83,680 J
P
 t 

 279 s  4.65 min
J
t
P
300
s
Q  mcT  250 g  1
The second way to solve this problem is much more elegant so let’s look at it because this is how
physicists think… Do all the algebra first by substituting for the energy part of the power
definition the heat expression.
P
Energy Q mcT


t
t
t
Solve the last equation for t
cal
 80  C

mcT
g C
t 

 279s  4.65 min
J  1cal 
P
300  
s  4.184 J 
250 g  1
Either way, the aquarium would take 4.65 minutes to warm up with this heater.
11. A 500.0 g pot of water at room temperature (20 degrees C) is placed on a stove. How
much heat is required to change this water to steam at 100 degrees C?
This is a problem that involves heating through a phase change (liquid to gas). So we
must be prepared to use the appropriate latent heat term to account for the heating during the
phase change when the temperature is not changing. We will have two heat terms in this
** SOLUTIONS Jaquin’s Sections **
SCI111 Homework Assignment #5
(20 Points Total)
problem: one to warm the water to 100 degrees C and the second to boil the water into steam at a
constant 100 degrees C.
cal
 80  C  40,000cal

g C
cal
Q2  mLv  500 g  540
 270,000cal
g
QTotal  Q1  Q2  40,000cal  270,000cal  310,000cal  310kcal
Q1  mcT  500 g  1
The heat need is 310 kcal.
13. Lead is a soft dense metal with a specific heat of 0.028 kcal/kgC, a melting point of
328 degrees C, and a heat of fusion of 5.5 kcal/kg. How much heat must be provided to
melt a 250.0 kg sample of lead with a temperature of 20 degrees C?
This is a problem that involves heating through a phase change (solid to liquid). So we must be
prepared to use the appropriate latent heat term to account for the heating during the phase
change when the temperature is not changing. We will have two heat terms in this problem: one
to warm the lead to 328 degrees C and the second to melt the lead into a liquid at a constant 328
degrees C.
kcal
 308 C  2,156kcal

kg  C
kcal
Q2  mLF  250kg  5.5
 1,375kcal
kg
QTotal  Q1  Q2  2,156kcal  1,375kcal  3,531kcal
Q1  mcT  250kg  0.028
The heat need is 3,531 kcal.
In a long paragraph, carefully explain why the saucy end of a just-removed-from-the-oven
piece of pizza will burn your mouth more than the crusty end. Highlight the concepts of
heat capacity, specific heat and the difference between temperature and heat.
1) The following paragraphs are from http://www.mevio.com/episode/87195/the-science-ofpizza
** SOLUTIONS Jaquin’s Sections **
SCI111 Homework Assignment #5
(20 Points Total)
If you have ever eaten pizza, you have probably burned your mouth on the cheese. If the crust of the pizza is cool enough for
you to hold, why is the cheese still hot enough to burn you? To find out, you will need: a nice, hot pizza.
Cut yourself a nice, hot slice of pizza, but do not take a bite. Instead, touch the crust. It will be very warm, but it should not be hot
enough to burn you. Do not touch the sauce or the cheese, because they will be hot enough to cause a burn. Why?
It has to do with a property known as specific heat. That is the amount of heat energy that a substance has to absorb to raise its
temperature. Some substances, like the crust of the pizza, have a low specific heat. They do not have to absorb much heat energy
to raise their temperature. That means that they do not have to lose much heat energy to cool off. The pizza crust was able to
transfer enough heat to your fingers to cool it, without transferring enough to burn your fingers.
On the other hand, the sauce and cheese have a high specific heat. It takes quite a bit of heat energy to warm them up. If you
tested the temperature while the pizza was cooking, you would find out that in a 450 degree oven, the crust quickly reaches 450
degrees, while the sauce and cheese take much longer. Since they have to absorb a lot of heat to get to that temperature, they
have to get rid of the same amount of heat energy to cool down. If you touch the sauce too soon, quite a bit of that heat energy is
transferred to your skin, giving you a burn.
So the next time you have pizza, let it cool a bit before you take your first bite.
2) The following paragraphs are from http://cooking.stackexchange.com/questions/6287/whydo-tomatoes-get-so-hot
All cooked food gets hot, and everything in any given dish will have the same temperature {*}. The tomatoes don't get hotter
than the other ingredients. But they do have a tendency to burn more than certain other substances, so the question is
"Why?"?
You get burned when a portion of your flesh reaches a high enough temperature{+}. The food warms your tongue, lips, etc. by
heat conduction until either you move the food or your mouth parts and the food reach the same temperature (a condition
known as thermal equilibrium).
What that common temperature is depends on the amount of heat (i.e. thermal energy) in the system. Some of the factors that
come into play are:
How much (mass of) food there is.
How much (mass of) your mouth is involved (see below).
The initial temperature of the food.
The "specific heat" of both the food and your mouth parts, which is a property of each substance that appears as a
coefficient in the thermal equilibrium equation. Water has a (very!) high specific heat, so watery foods, if they are at a high
final temperature, tend to burn you more easily. There is an added complication for the extra heat needed to establish a
phase change (i.e. melt solids or vaporize liquids) called the heat of fusion or heat of vaporization. Again water has a high
value for both of these numbers.
How fast the common temperature is reached depends on the area of contact between the food and the mouth.
Another coefficient called the thermal conductivity. This one is complicated, but liquids tend to have a high thermal
conductivity and solids less so. This is where soups, sauces, and melted cheese really get you. Note that your mouth
parts has a pretty low thermal conductivity, so you only get to count the surface layers in finding the equilibrium
temperature. Sorry.
Some consequences of all this:

This is why you can peel the aluminum foil off of a pan that has just come out of a 400 degree (F) oven without
trouble, but if you get your hand stuck in the steam plume (which is only around 212 degrees F) you get
scalded: Aluminum has a low heat capacity, and steam has a (very, very!) high one.

Small bites help in two ways: less total heat means a lower common temperature, and may allow you to move
the food around in your mouth, reducing the temperature of any one part.
Some foods are just dangerous this way. You know what they are from experience: steam, hot soups and sauces, melted
cheese, etc.
{*} Well, sort of. But take that as true for any particular region of any particular dish.
{+} What temperature is that? Good question. Maybe there is a medical professional around, 'cause I don't know. I'd
guess around 140--150 degrees F (call it 60--65 degrees C), but don't quote me.
** SOLUTIONS Jaquin’s Sections **
SCI111 Homework Assignment #5
(20 Points Total)
3) According to Wikipedia…Water has the second highest specific heat capacity of all known
substances.
4) Let’s try to quantify the problem.
A)
Let’s assume the average piece of pizza has the dimensions illustrated in the figure below.
15 cm
20 cm
2 cm
4 cm
Let’s for simplicity assume the sauce/cheese end of the pizza has the density and specific heat of water.
Let the crust material have half the density and half the specific heat of water.
B) The volume of the sauce/cheese end of the pizza will be given by
V
1
1
bh  t  15cm  20cm 1cm  150cm 3
2
2
of sauce and 150cm3 of crust. The mass of sauce and crust are about (m=∙V) are 150 g and 75 g
respectively.
C) The heat released from the saucy end of the pizza as it cools from 450 F/232 C (oven temperature) to 98
F/37 C (body temperature) is
QTotal  QSauce  QCrust
 mSaucecSauceT  mCrust cCrust T
cal
cal
195 C  75 g  0.5  195 C

g C
g C
 29.25kcal  14.6kcal
 150 g 1.0 
 43.8kcal
The crusty end will have volume of 4cm∙4cm∙15cm = 240 cm3 and mass of 120 g. The heat released from
the crusty end will be 11.7 kcal.
D) Let’s assume that one bite of crust contains 40 g of crust and one equal volume bite of the saucy end
contains 80 g. The heat released into your mouth for one bite of crust will be
** SOLUTIONS Jaquin’s Sections **
SCI111 Homework Assignment #5
(20 Points Total)
11.7kcal 
40 g
 3.9kcal
120 g
The heat released into your mouth for one bite of the saucy end will be
43.8kcal 
80 g
 15.6kcal
225 g
E) So we see that the saucy end of a pizza has almost four times heat content per bite compared to the crusty
end. Ouch! It is a combination of the high specific heat and higher density of the saucy end that delivers
more heat to your mouth than the crust of a hot piece of pizza fresh out of the oven.