ANSWER AND DISCUSSION
PROBLEM #1, STATES OF MATTER
1. How much heat is needed to warm 25 grams of water from 10.ºC to 20.ºC?
(The decimals after the 10 and 20
have just been placed there to indicate that the numbers are accurate to two
significant digits.) See the graph below for a picture of what is happening in these problems.
Heat is added and the temperature of the water increases. That is all that
is happening in this problem. Leg "C" of the graph between 10.ºC and 20.ºC is the
only part of the graph that is needed. There is only a straight line graph
between these two points, meaning that this is the only thing happening and
that the entire process can be described by one equation, Q = m c Δ T.
The sloped line between the two points indicates that there is a change of heat with a change of temperature.
This problem is likely best done with W5P method because it asks for an
amount of heat when you are given two temperatures, a mass, and a type of
material. The measurements such as the specific heat, the heat of vaporization,
or the heat of fusion, the properties of those materials, are all available
if you know the type of material. In the special case of water, you should
know all of those property quantities. The English
units are not commonly used except in some industrial applications, such as
air conditioners being rated in BTU's.
GIVEN:
T_{1} =
10.ºC, T_{2} =
20.ºC, m = 25 g, water is in
the liquid state at the temperatures in this problem, so
c = 1.00 cal/gdeg.
FIND: a heat,
Q The question asks for a heat in calories.
FORMULA:
Q = m c Δ T or
Q = m c ( T2
 T1 )
You know two temperatures, a mass, and a specific heat.
You are asked to find a heat. The formula comes from the formula section of Units in Chemtutor.
SOLVE:
The formula is already solved for heat as it is memorized. Substitute the known values from
GIVEN into the working equation, simplify, and do the math. No DA is
needed to change units, but carefully handle the units. The units can help find any math
problems you may have.
WHAT YOU WRITE DOWN SHOULD LOOK LIKE THIS:
GIVEN:
T_{1} = 10.ºC,
T_{2} = 20.ºC,
m = 25 g, c= 1.00
cal/gdeg
FIND: a heat,
Q
The calculator math is: (20  10) x 25 =
ANSWER:
Calculate the numerical answer and show your cancellation of the units. The
raw answer does not need to be changed into scientific notation or shortened
to any fewer significant digits.
Back to Problems in States of Matter.
Back to Problems
in States of Matter.
ANSWER AND DISCUSSION
PROBLEM #2, STATES OF MATTER
2. How much heat is needed to warm 25 grams of water (H_{
2}O) from 10.ºC to 20.ºC?
This problem involves heating ice to the melting point, a change of state (ice to water), and
heating as water. There are three legs to the temperature change of water
from 10ºC to 20ºC labelled "A" as the first leg, the warming of
ice to 0ºC, "B" as the melting of ice to water at 0ºC, and "C" as
the warming of water from 0ºC to 20ºC. The two warming stages, as
ice and as water, have the same formula, Q = m c ΔT, relating the dimensions, but the numbers
may be different. The specific heat of ice, "c_{i
}," is 1/2 calorie per gramdegree whereas the specific
heat of liquid water,
"c_{w}," is one calorie per gram degree.
The mass of the H2O is the same in this case. The
formula for the melting leg of the change is, Q = m H_{f
}, where Q is the heat, m is the same mass as in the other
legs, and H_{f} is the heat of fusion of
water. The total formula for finding the amount of heat for this temperature
and state change is the sum (total) of the three formulas as solved for Q.
The graph above shows the three legs of the change of
water from ice at 10ºC to 20ºC, and the problem solution below
labels the legs, i for ice, f for fusion, and w for water.
GIVEN:
T_{1i} = 10ºC,
T_{2i} = 0ºC,
m = 25 g,
c_{i} = 1 cal/gramdegree,
T_{1w} = 0ºC,
T_{2w} = 20ºC,
m_{w} =
25 g, c_{w
} = 1 cal/gramdegree, and H
_{f} = 80 cal/g
FIND: a heat, Q
The question asks for a heat in calories.
FORMULAS:
Q = m c ΔT and Q = m H_{f}
The formula for temperature increase with heat application
(Q = m c ΔT) will be used for the two legs A and
C, the increase in temperature of the H_{2}O
as ice and as water. The formula for the energy used to melt the ice, Q = m H_{f}, must also be used. The total amount of heat needed must be the heat needed to warm the ice to 0 ºC, change the solid water (ice) to liquid water, still at 0 ºC, and warm the water to 20 ºC.
SOLVE:
The formulas are already solved for heat as they are memorized. The heat of
the legs must be added together to find the total heat, Q_{T},
or, because the mass is the same in all three equations,
the equations may be rearranged as:
Q_{T} = m
c_{i} ΔT_{i} +
m H_{f} + m c_{w}
ΔT_{w}
or
Q_{T} = m [
(c_{i} ΔT_{i}) +
H_{f} +
(c_{w} ΔT_{w})]
Where Q = heat in calories, c = specific heat, ΔT =
the change in temperature, H_{f} = the heat
of fusion, m = the mass, and the subscripts of "i" and "w" are for ice and water
respectively. Substitute the known values from GIVEN
into the working equation, simplify, and do the math.
WHAT YOU WRITE DOWN SHOULD LOOK LIKE THIS:
GIVEN:
T_{1i} = 10ºC,
T_{2i} = 0ºC,
m = 25 g,
c_{i} = 1 cal/gramdegree.
T_{1w} = 0ºC,
T_{2w} = 20ºC,
m_{w} = 25 g,
c_{w
}= 1 cal/gramdegree, and H_{f} = 80 cal/g.
FIND: the total
heat, Q_{T}
FORMULAS:
Q_{f} = m H_{f}
Q_{i}
= m c_{i}
( T_{2i}
 T_{1i}
)
Q_{w}
= m c_{w}
( T_{2w}
 T_{1w}
) Q_{T}
= Q_{i}
+ Q_{f}
+ Q_{w}
SOLVED:
Q_{T}
= m c_{i}
( T_{2i}
 T_{1i}
) + m H_{f}
+ m c_{w}
( T_{2w}
 T_{1w}
)
alternate:
Q_{T}
= m [c_{i}
( T_{2i}
 T_{1i}
) + H_{f}
+ c_{w}
( T_{2w}
 T_{1w}
)]
ANSWER:
Calculate the numerical answer and show your cancellation of the units.
It is fascinating that melting of ice takes a lot more energy in the form
of heat than a thirty degree change in temperature. Even so, the 80 cal/gram heat of
fusion is still a lot smaller than the 540 cal/gram heat of vaporization.
Back to Problems in States of Matter.
ANSWER AND DISCUSSION
PROBLEM #3, STATES OF MATTER
3. What is the specific heat of copper metal if 200 cal increases the
temperature of 40.9 g of it from 21ºC to 73ºC?
With other temperatures, heats of phase
change, and specific heats that fit the properties of copper, the graph
for water is similar to the graph for copper metal. At the temperatures in this
problem, copper is on the first leg (Leg "A") as a solid. There is
no change in state. The formula must be Q = m c ΔT.
The property quantities such as the value of the specific heat of copper are
not commonly memorized measurements. Your instructor will not likely ask you
to know such quantities. If you need to use such a number in your calculations and the information is not available to you in a problem, you should consult tables with these properties such as the table of heat properties of selected materials.
GIVEN:
Q = 200 cal,
T_{1} =
21ºC,
T_{2} =
73ºC,
m= 40.9 g,
The material is copper in
the solid state at the temperatures in this problem.
FIND: a specific heat, c
The question asks for a specific heat in calories per gram.
FORMULA: Q = m c ΔT or Q = m c ( T2  T1
)
You know two temperatures, a mass, and a heat.
You are asked to find a specific heat. The formula comes from the formula section of Units in Chemtutor.
SOLVE:
The formula must be solved for specific heat from the memorized formula. Substitute the known values from
GIVEN into the working equation, simplify, and do the math. No DA is
needed.
WHAT YOU WRITE DOWN SHOULD LOOK LIKE THIS:
GIVEN:
Q = 200 cal,
T_{1} = 21ºC,
T_{2} =
73ºC, m = 40.9 g, The material is copper as a solid.
FIND: a specific heat,
c
FORMULA:
Q = m c ΔT
The calculator math is: 200 ÷ 40.9 ÷ 52 =
ANSWER:
Calculate the numerical answer and show your cancellation of the units. The
raw answer does not need to be changed into scientific notation but does
need to be shortened to two significant digits. How does this answer match
the value for the specific heat of copper in the table?
Back to Problems in States of Matter.
ANSWER AND DISCUSSION
PROBLEM #4, STATES OF MATTER
4. How much heat is needed to increase the temperature of 12 grams of gold from 20 ºC to 95 ºC?
You know enough about the properties of gold to see that it would be a
solid in all the temperatures in this example. Q = m c ΔT is the equation that describes this change. You are given a mass,
two temperatures, and the material. You are asked for an amount of heat.
Since the type of material is available, it is assumed that you have access
to a table that would give you the specific heat,
a property of the material.
GIVEN:
T_{1} =
20ºC,
T_{2} =
95ºC,
m= 12 g, and the material is gold in
the solid state, so from the table,
c = 0.032 cal/gramdegree.
FIND: a heat,
Q
The question asks for a heat in calories.
FORMULA:
Q = m c ΔT or Q = m c ( T2  T1
)
You know two temperatures, a mass, and a specific heat.
You are asked to find a heat. The formula comes from the formula section of Units in Chemtutor.
SOLVE:
The formula is already solved for heat as it is memorized. Substitute the known values from
GIVEN into the working equation, simplify, and do the math. No DA is
needed, but carefully handle the units. The units can help find any math
problems you may have.
WHAT YOU WRITE DOWN SHOULD LOOK LIKE THIS:
GIVEN:
T_{1} =
20ºC, T_{2} =
95ºC, m = 12 g, c = 0.032 cal/gramdegree.
FIND: a heat,
Q
The calculator math is: 12 x 0.032 x 75 =
ANSWER:
Calculate the numerical answer and show your cancellation of the units. The
raw answer does not need to be changed into scientific notation, but it needs to be shortened
to two significant digits. It is fascinating that with only a very small
amount of added heat, the gold increases temperature so much compared to water.
Back to Problems in States of Matter.
ANSWER AND DISCUSSION
PROBLEM #5, STATES OF MATTER
5. What is the temperature of 40g of water at 45ºC
added to 60g of water at 95ºC?
This is a neat little question about the
second law of thermodynamics. The
new temperature of the resulting material from a mix will all have the same
temperature and that temperature will be the weighted average of the masses
and temperatures and specific heats of the contributing materials. Since there
is no change in phase, the problem goes without that complicating factor. We
can treat the temperatures as absolute numbers, even though they are not. The
first law of thermodynamics, that energy can be traded around but not lost,
is the basis for the equation for this happening. The equations are:
Q_{1} + Q_{2
}= Q_{T} and
c_{T} = c_{1
}= c_{2} and
m_{1} + m_{2} =
m_{T}, so
[(m_{1}T_{1
}) + (m_{2}T_{2})
/(m_{1} + m_{2})
] = T_{T }.
If these were different materials added together, we
would have to take into account the specific heat of each of them, but
both materials here are water in the liquid state.
For many of these problems there will be a material that gains heat
and a material that loses heat. Strictly speaking, a positive value for Q
should indicate a gain of heat (since T_{2} 
T_{1} is a positive number when there is
an increase in temperature) or a greater amount of the looser phase, and a
negative value for Q when there is a temperature loss or greater amount of the
more tightly  bound phase. On the other hand, if we consider "heat lost" as a
positive amount we can more easily construct formulas on the basis of,
Q_{gained} = Q_{lost}
rather than the slightly more inconvenient Q_{gained} =
 (Q_{lost}), where Q_{lost}
is a negative number.
GIVEN:
T_{1} =
45ºC
T_{2} =
95ºC
m_{1} = 40 g
m_{2} = 60 g
m_{1}
+ m_{2} =
m_{T} = 100 g c
_{1}
= c_{2} = c
_{T} = 1.00 cal/gramdegree
FORMULAS: Q = m c ΔT or Q = m c ( T2  T1
) and Q_{1}
+ Q_{2}
= Q_{T}, the equation of the first law of thermodynamics, and m_{1}
+ m_{2} =
m_{T}, the adding of
the masses of water.
SOLVE:
With the several formulas, T_{T} must be
solved for.
WHAT YOU WRITE DOWN SHOULD LOOK LIKE THIS:
GIVEN:
T_{1} =
45ºC, T_{2} =
95ºC, m_{1} = 40 g,
m_{2} = 60 g,
FIND:
T_{T}, the final temperature.
The calculator math is: 40 x 45 = STO 60 x 95 + RCL = ÷ 100 =
The "STO" is a storage function and "RCL" is a reclaim function. These
may be "M+" for, "add to memory," and "m" for, "retrieve from memory," or some
other such set of buttons, depending on your calculator.
ANSWER:
Calculate the numerical answer and show your cancellation of the units. The
raw answer does not need to be changed into scientific notation or shortened
to any fewer significant digits.
Back to Problems in States of Matter.
ANSWER AND DISCUSSION
PROBLEM #6, STATES OF MATTER
6. What mass of live steam at 100ºC is needed to heat
27.5 kg of water from 20ºC to 100ºC?
This problem is a good example of the idea that the formulas in this section
are easy, but it often takes some skill and insight to construct the set of
formulas in the correct way to reflect what is going on in the situation. See
the graph for a good mental image of what is happening in
this problem. The two materials, the steam at 100ºC is added to the water
at 20ºC. The amount of heat gained by the water to get to be water at
100ºC has to be the same amount of heat lost by the steam to make the same
water at 100ºC, the same temperature and phase as the other water.
Let's call the amount of heat gained by the 20ºC water
Q_{c} because it is a heat, the heat
on leg "c." Similarly, let's call the amount of heat lost by the steam
Q_{d} because it involves leg "d." The
formula is: Q_{c} = Q_{d}.
So m_{c} c_{c} ΔT_{c} =
m_{d} H_{V}.
We know m_{c}, c_{c},
ΔT_{c}, and
H_{V}, and we are looking for
m_{d}
WHAT YOU WRITE DOWN SHOULD LOOK LIKE THIS:
GIVEN:
T_{1} =
20ºC,
T_{2} =
100ºC, m_{c} = 27.5 Kg ( = 2.75 E4 g)
From the properties of water table, c_{c} = 1.00 cal/gramdegree,
H_{V} = 540 cal/g
FIND: a mass,
m_{d}
The calculator math is: 27.5 X 80 ÷ 540 =
ANSWER:
Calculate the numerical answer and show your cancellation of the units. The
raw answer does not need to be changed into scientific notation or shortened
to any fewer significant digits. It is fascinating that with only a very small
amount of added heat the gold increases temperature so much compared to water.
Back to Problems in States of Matter.
ANSWER AND DISCUSSION
PROBLEM #7, STATES OF MATTER
7. A 1.1 kg (1100 gram) iron horseshoe at 1300 °C is dunked into a
bucket containing 12.8 kg of water at 20 °C. Assume that no steam was
lost. (All of the steam leaves its heat of vaporization and its mass to the
water in the bucket.) The horseshoe is taken out at 100 °C, just after
it has quit boiling the water immediately around it. What is the new temperature of the
bucket of water?
Here the horseshoe loses Q = m c ΔT heat to
the water and the water gains all that heat to increase the temperature of the
water. The first important process is to see that happening, and the second important
process is to carefully label everything so we can analyze it. Let's call the
amount of heat lost by the iron Q_{i}
and the amount of heat gained by the water Q_{w}. Therefore,
Q_{i} = Q_{w}.
GIVEN:
T_{1w} =
20ºC
T_{1i} =
1300ºC
T_{2i} = 100ºC
m_{i} = 1.1 Kg
m_{w} = 12.8 Kg
The specific heats,
c_{i} = 0.12 cal/gdeg and c_{w} = 1.00 cal/gdeg
come from the table of information on properties of materials.
FIND:
the final temperature of water, T
_{2w}
FORMULA:
Q = m c ΔT or Q = m c ( T_{
2}  T_{1}
)
What is more important than the memorized equations is how
to put them to use. Here for the iron horse shoe,
Q_{i} =
m_{i}
c_{i}
ΔT_{i}
or
Q_{i} =
m_{i}
c_{i}
( T_{2i}
 T_{1i} )
and for water,
Q_{w} =
m_{w}
c_{w}
ΔT_{w}
or
Q_{w} =
m_{w}
c_{w}
( T_{2w}
 T_{1w} )
We know the mass of iron, the specific heat of iron, the initial and final
temperatures of iron. We know the mass of water, the specific heat of water, and
the initial temperature of water. The formulas come from the formula section of Units in Chemtutor.
SOLVE:
Solve for the final temperature of the water. Substitute the known values from
GIVEN into the working equation, simplify, and do the math.
WHAT YOU WRITE DOWN SHOULD LOOK LIKE THIS:
GIVEN:
T_{1w} =
20ºC, T_{1i} =
1300ºC, T_{2i} =
100ºC, m_{i} = 1.1 Kg,
m_{w} = 12.8 Kg,
c_{i} = 0.12 cal/gdeg
, and c_{w} = 1.00 cal/gdeg.
FIND: the
final temperature of water, T_{2w}
The calculator math is: 1.1 x 0.12 x 1200 = STO 12.8 x 20 = + RCL = ÷ 12.8 =
ANSWER:
Calculate the numerical answer and show your cancellation of the units. The
raw answer does not need to be changed into scientific notation, but it must be
shortened to two significant digits.
Look at that problem again. Why is there such a tiny change in temperature of the water compared to the large change in iron temperature? Well, there is a very large difference in the specific heat ("heat capacity") of water versus iron, and there is a large difference in the masses of the two materials.
Back to Problems in States of Matter.

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