hello and welcome to the chemistry
solution this tutorial is on heat
capacity and specific heat heat capacity
is the amount of heat required to raise
the temperature of an object by one
degree Celsius the difference between
heat capacity and specific heat is that
heat capacity is an extensive property
and it's dependent on the amount of a
substance that you have specific heat in
contrast is an intensive property and is
independent of the sample size specific
heat is the amount of heat required to
raise the temperature of one gram of a
material by one degree Celsius
another way of saying extensive or
intensive properties is to say extrinsic
or intrinsic we know that Q is energy in
the form of heat and Q is equal to the
specific heat of a substance times the
grams of that substance times the
temperature change let's go ahead and
look at an example how many kilojoules
of heat are required to raise the
temperature of 365 grams of water from
68 degrees Celsius to 90 degrees Celsius
given that the specific heat of water is
4.18 joules per gram degree Celsius we
know that Q is equal to the specific
heat times the grams of the substance
times the change in temperature so Q is
going to be equal to our specific heat
so the specific heat is 4.18 joules for
every gram degree Celsius this means
that 4.18 joules of energy are required
to raise one gram of water one degree
Celsius we know that we have 365 grams
of water because the specific heat is
given to us in joules per gram degree
Celsius and our sample size is given to
us in grams this allows us to cancel out
units of grams we know that delta T are
the change in temperature is equal to 22
degrees Celsius and we calculate delta T
by
taking our final temperature minus our
initial temperature so our final
temperature is ninety degrees Celsius
minus the temperature we started at or
68 degrees Celsius this gives us a delta
T value of 22 degrees Celsius and this
is something important to note some
times specific heat values will be given
to you in joules per gram Kelvin and
although the numerical values for a
temperature given in degrees Celsius or
Kelvin are different delta T in degrees
Celsius and Delta Kelvin for two given
temperatures will be the same and so
let's take a look at that remember that
Kelvin is equal to a temperature in
degree Celsius plus 273 so the absolute
values of Kelvin and degree Celsius are
different numbers but a change in one
degree Celsius is equal to a change in
one Kelvin and so if my final
temperature was 90 degrees Celsius that
would equal a temperature of 363 Kelvin
and my initial temperature was 68
degrees Celsius that would be equal to
341 Kelvin if I now look at Delta T I
can see that delta T is 22 degrees
Celsius and is also 22 Kelvin so for
problems involving delta T as long as
your values are given to you in degree
Celsius or Kelvin you do not need to
convert these temperatures in order to
cancel with your given specific heat
value if your specific heat value is
given to you in different units and
remember that's only if you're working
with Celsius and Kelvin this doesn't
apply to any temperature is given in
Fahrenheit but typically when you're
working with problems in the sciences
you're working on either the Celsius or
the Kelvin scale going back to our
problem we can cancel degrees Celsius
and we come up with an answer of 33600
joules but our problem is asking us how
many kilojoules of heat are required so
we need to convert Joules to kilojoules
remember that one
thousand joules is equal to one
kilojoule this allows us to cancel out
units of joules and come up with a final
answer of thirty three point six
kilojoules let's look at another example
what is the specific heat of aluminum if
2230 joules are required to raise the
temperature of a 45 grams sample from 30
degrees Celsius to 85 degrees Celsius
we'll go back to using our same equation
although this time we're going to solve
for the specific heat so we can
rearrange that equation to tell us that
the specific heat is equal to Q divided
by the mass of the sample divided by
delta T we know that Q is equal to 2230
joules as given to us in the problem we
divide that by the mass of the sample 45
grams and we also divide Q by the change
in temperature delta T which is 55
degrees Celsius and you'll notice that
none of our units cancel and that's all
right because specific heat values are
normally given to us in joules per gram
degree Celsius and when you punch these
numbers into your calculator you should
come up with point nine zero zero joules
per gram degree Celsius thanks for
watching the chemistry solution we hope
you enjoyed this tutorial
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