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Pirate Chemistry 2009
Problem Solving with Heat
Heat is quite a complex concept. Heat can be effected by how much of the substance there is,
what temperature the substance is at, and what the substance is. We need a unit define and
quantify heat. It is a unit that you are very familiar with already; the calorie.
Calories
A calorie is a unit of energy, of heat. With every substance having a different specific heat capacity, it can be difficult to come up with a unit that describes all energy. It was decided to pick
a common substance to act as a standard against which all other substances would be measured.
The common substance picked was water. We know that mass, specific heat, and temperature
all effect the energy of a substance so to keep the numbers very simple:
a calorie was defined as the amount of energy needed to raise 1 grams of water by 1 oC
Do you notice that all three aspects of heat are included; mass, temperature, and identity? Thus,
the specific heat capacity of water (as indicated in the previous section) is defined as 1 cal/goC.
So if 1 calorie of heat is applied to 1 gram of water at 20 oC, the water increases to 21 oC. If 10
calories of heat are applied to 1 gram of water at 20 oC, the water increases to 30 oC. If 10 calories of heat are applied to 10 grams of water at 20 oC, the water increase to 21 oC! Let’s put all
this into a simple equation which includes all of our variables:
Q = mc Tf—Ti)
Q = mc T)
Heat
Mass
Specific heat
capacity
Change in
temperature
Final temperature minus the inititial temperature
The heat quantity is Q, the mass is m, the specific heat capacity is c, and the final temperature is
Tf while the initial temperature is Ti.. Another way to represent this equation is by combining
Tf and Ti into one variable T, which is the change in the temperature; the same thing as Tf-Ti.
Lets look at some examples of typical problems you will see:
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Pirate Chemistry 2009
Example 1:
How much heat (in calories) is needed to raise 20 grams of water from 5 oC to 40 oC?
Q = mc(Tf—Ti)
Q=x
m = 20 grams
c = 1 cal/goC (this is the specific heat of water)
Tf = 40 oC
Ti = 5
x = (20 g) (1 cal/goC)(40 -5 0C) = 700 cal
Example 2:
How much heat (in calories) is needed to raise 140 grams of water from 20 oC to 25 oC?
Q = mc(Tf—Ti)
Q=x
m = 140 grams
c = 1 cal/goC (this is the specific heat of water)
Tf = 25 oC
Ti = 20 oC
x = (140 g) (1 cal/goC)(25-20 0) = 700 cal
Please note that the amount of calories is the same in both problems.
In example 1 we have 20 grams being changed by 35 oC while in example 2 we have much more
water (140 grams) being changed by a much smaller amount (5oC).
Example 3:
How much heat (in calories) is needed to raise 250 grams of water from 80 oC to 87 oC?
Q = mc(Tf—Ti)
Q=x
m = 250 grams
c = 1 cal/goC (this is the specific heat of water)
Tf = 87 oC
Ti = 80 oC
x = (250 g) (1 cal/goC)(87—80 0C) = 1750 cal
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it will be.
Pirate Chemistry 2009
Example 4:
How many calories are needed to raise 50 grams of iron from 55 oC to 200 oC? The specific
heat capacity of iron is 0.11 cal/goC.
Q = mc(Tf—Ti)
Q=x
m = 50 grams
c = 0.11 cal/goC
Tf = 200 oC
Ti = 55 oC
x = (50 g) (0.11 cal/goC)(200– 55 0C) = 797.5 cal
Example 5:
How many grams of aluminum can be heated from 90 oC to 120 oC if 500 calories are applied?
The specific heat of aluminum is 0.21 cal/goC.
Q = mc(Tf—Ti)
Q = 500 calories
m = x grams
c = 0.21 cal/goC
Tf = 120 oC
Ti = 90 oC
500 cal= (x) (0.21 cal/goC)(120—90 0C)
x = 79.4 grams
Example 6:
What is the specific heat capacity of a substance if 400 calories cause 25 grams of it to go from
60 oC to 190 oC?
Q = mc(Tf—Ti)
Q = 400 calories
m = 25 grams
c = x cal/goC
Tf = 190 oC
Ti = 60 oC
400 calories = (25 g) (x)(130 0C)
x = 0.123 cal/goC
All text copyright Chris Smith 2009. All pictures obtained from internet and are copyright of their owners but assumed to be public accessible. If you are the owner of a picture and want it removed, email [email protected], and
it will be.
Pirate Chemistry 2009
Example 7:
What is the final temperature if 500 calories are applied to 40 grams of copper at 20 oC? The
specific heat capacity of copper is 0.092 cal/goC.
Q = mc(Tf—Ti)
Q = 500
m = 40 grams
c = 0.092 cal/goC
Tf = x
Ti = 20 oC
500 cal = (40 g) (0.092 cal/goC)(x– 20 0C)
x = 156 oC
Example 8:
What was the initial temperature if 250 calories were applied to 100 grams of gold and the final
temperature of the gold was 175 oC? The specific heat capacity of gold is 0.031 cal/goC.
Q = mc(Tf—Ti)
Q = 250 calories
m = 100 grams
c = 0.031 cal/goC
Tf = 175 oC
Ti = x oC
250 cal= (100 g) (0.031 cal/goC)(175—x 0C)
x = 94 oC
Example 9:
What is the change in the temperature if 75 calories are applied to 10 grams of water?
Q = mc(Tf—Ti)
Q = 75 calories
m = 10 grams
c = 1 cal/goC
Tf –Ti = x
(note that in this problem we want the change so both Tf—Ti are included as x)
75 calories = (10 g) (1 cal/goC)(x)
x = 7.5 oC
All text copyright Chris Smith 2009. All pictures obtained from internet and are copyright of their owners but assumed to be public accessible. If you are the owner of a picture and want it removed, email [email protected], and
it will be.
Pirate Chemistry 2009
Questions
Specific Heat Capacities
Substance
Specific Heat Capacity (cal/goC)
Water
1.0
Ice
0.49
Copper
0.092
Gold
0.031
Iron
0.11
Aluminum
0.21
1. How many calories would it take to raise the temperature of 200 grams of water from 5 oC
to 85 oC?
2. How many calories would problem number 1 be if it was aluminum instead of water?
3. How many grams of copper could be heated from 20 oC to 75 oC if 1200 calories are applied to it?
4. How many grams of iron could be heated from 15 oC to 300 oC if 8000 calories are applied
to it?
5. What is the specific heat capacity of a substance if 750 calories caused 100 grams of it to go
from 90 oC to 135 oC?
6. When 400 calories are applied to 20 grams of a substance, it goes from 62 oC to 82 oC.
Which of the substances in the table above is this?
7. What is the final temperature if 400 calories are applied to 225 grams of iron at 40 oC?
8. What would the final temperature be if 500 calories are applied to 150 grams of ice at –90
o
C?
9. What is the temperature change if 50 calories are applied to 4 grams of water?
10. What would the temperature change by if a 90 gram piece of hot iron cooled by losing 200
calories.
We typically think of calories as the energy from food. However,
food calories are actually Calories with a capital C. This small
change is very important as a Calorie is equal to 1000 calories or 1
kilocalorie. This nutrition fact shows that an egg can give 70 Calories which would be 70,000 calories. This is enough energy to
raise 70,000 grams of water by 1 oC!
A typical diet is supposed to have about 2000 Calories or 2,000,000
calories which seems like a lot. Keep in mind, though, that our
bodies have to keep us at about 31 oC every minute of the day and a
typical person weighs around 175 pounds or around 80,000 grams.
Also, our bodies are not perfect in turning every bit of that energy
into useful, productive outcomes.
All text copyright Chris Smith 2009. All pictures obtained from internet and are copyright of their owners but assumed to be public accessible. If you are the owner of a picture and want it removed, email [email protected], and
it will be.