The specific heat capacity of an unknown liquid is #0.32# #"J"/("kgK")#. The density of the liquid is #0.0321# #"g/mL"#. If a chemist applies #243# #"J"# of heat to #300# #"mL"# of this liquid starting at #27.1 ^@# #C#, what is the final temperature?
Answer is #80000^@"C"#
Thanks in advance
Answer is
Thanks in advance
2 Answers
Given,
Now, recall,
Hence,
is the final temperature of the solution, given your data.
80000 C
Explanation:
So we start with the equation q=mc
Since
We're given the initial temp. (
However, we're not given the final temp. (
This means we first have to find,
We do this by taking the equation q=mc
It should now look like
q/mc =
So we have our rearranged equation, but we first need to do some conversions before we plug anything in.
The
Since we are told that
Now we have to convert
Now that is done, we can plug our values into the equation
q/mc =
Our
Our
Our
So now with terms plugged in, it should look like this
(
Our Joules and kg will cancel out, and leave us with
(
Now multiply the
Now divide the
So
Since our final answer is in
We do this by subtracting
So now our
Then we take our equation,
Add
Your final answer should be
But with it rounded, it's