okay let's practice measuring specific heat. so the question says a fifty nine
point seven gram piece of metal is transferred from boiling water so it
started out in boiling water and then we're gonna put it in 60 point zero
milliliters of water that is initially at twenty two point zero degrees Celsius
if the final temperature of the system so the metal in the cold water ends up
at twenty-eight point five degrees Celsius what is the specific heat of the
metal. so here's all the information, we've got that hot metal that started in
boiling water and we know what temperature boiling water has to be and
we put it in that cold water and it goes from twenty two to twenty eight point
five and we want to find the specific heat. so if this is daunting go ahead and
check out my tutorial on thermo chemistry and specific heat and heat
transfer and that will outline what you need to do for this kind of problem and
when you're ready give it a try.
okay so let's split this up into two parts. first
we want to see what happened with the water and then later we'll see what
happened with the metal. so looking at the water let's set up our equation so
we know that this equation is the one that we're gonna want to use. so q or
heat is equal to c or specific heat times mass times delta T or the change
in temperature. so the heat absorbed by the water is going to be equal to the
specific heat of water times the mass of water times the change in temperature
that the water experienced. and the change in temperature of course is Tf
minus Ti, or the final temperature minus the initial temperature. so plugging
things in q is going to be equal to the specific heat of water which is four
point one eight six joules per gram degrees Celsius, and that's a tabulated
value we would just get that out of the textbook or online, that is
just known data. times the mass of the water and we had 60 milliliters of water
and water is one gram per milliliter so 60 milliliters is the same thing as 60
grams so we can just call that 60 grams of water. and then the change in
temperature or T final minus T initial. remember that the water ended up at 28
point five degrees Celsius that was the final temperature that's what it got
heated up to and then the initial temperature we subtract the initial
temperature from the final temperature it started out at twenty two point zero
degrees Celsius before we put the metal in there so that is the equation that
gives us how much heat was absorbed by the water and if we do the arithmetic we
get 1630 joules, that's to three sig figs. now the important thing to
understand is that the heat that was absorbed by the water is the heat that
was released by the metal so we have conservation of energy, right, the metal
got heated up and we put it in the water and it transferred energy into the water
so all of the heat that was absorbed by the water is precisely the heat that was
released by the metal or at least that's what we're going to assume in an
idealized situation so we can do another calculation with that same amount of
heat with the metal so let's take this equation again, q, or the heat released by
the metal is equal to c, the specific heat of the metal, this time times M the
mass of the metal, times delta T or T final minus T initial for the metal
rather than the water. so this is a new equation and we're looking at the metal
specifically so q we know it is going to be the opposite of the heat absorbed by
the water so instead of 1630 joules we have negative 1630 joules, that is the
heat that was released by the metal. now that is going to equal c which we will
leave as c because we do not know it yet that is what we're going to solve for so
we will leave that as c, times m which is fifty nine
point seven grams that was the mass of the metal that was given to us, times the
change in temperature for the metal. now for this we have to understand that the
T final is twenty eight point five degrees Celsius, remember that the system
as a whole equilibrated to twenty eight point five
degrees Celsius, that was the water with the metal in it so those together ended
up at twenty eight point five degrees Celsius all as a whole, so that we have
the same final temperature for the water and the metal. but from that we must
subtract the initial temperature of the metal and we're gonna call that a
hundred degrees Celsius because that was sitting in boiling water and water must
be a hundred degrees to boil so we're gonna call that a hundred degrees
Celsius. so that is our change in temperature for the metal so if we go
ahead and do the arithmetic we can solve for c that's going to be negative 1630
joules over the quantity fifty nine point seven grams times negative seventy
one point five degrees Celsius that's why it's important to do T final minus T
initial because we need a negative delta T in order to cancel out the negative q
value to give us a positive specific heat remember we're going to need a
positive number for specific heat a negative number wouldn't make any sense
so we need a negative delta T there so we plug all that into the calculator and
we get zero point three eight two joules per gram degree Celsius as the specific
heat of this metal.