Question

If steel has a heat capacity of 0.452 kj/kg/C. And you need to go raise the temperature by 1350 C. How many joules of heat would you need?

Answer 1

Hello @jellyfishdream
This question is very simple, but in order to solve the problem we need a vital information, as you might have noticed your self: We need to know the mass of the steel sample we want to raise the temperature of.

Answer 2

If we know the mass we can simply use the relation between the specific heat capacity:
\[\Large E=c \times m \times \Delta T\]

Answer 3

I know the Density is 7,874 Kg/m3

Answer 4

which numbers get plugged into where?

Answer 5

Well if we know the density, then we just get a new variable which is the volume,
\(Q\) is the energy added as heat, \(c\) is the specific heat capacity, \(\Delta T\) is the difference in temperature. \(\rho\) is the density and \(V\) the volume

Answer 6

Hmm then we just get (sorry I wrote wrong the first time)
\[\Large Q=c \times \Delta T \times \rho \times V\]

Answer 7

Q= 0.452 kj/kg/C x ΔT x 7,874 Kg/m3 x V

Answer 8

Where do I plug in the 1350 C?

Answer 9

in \(\Delta T\) as it is your increase in temperature.

Answer 10

But you still need a volume then.

Answer 11

As I said, we are lacking some information.

Answer 12

Furthermore the density is usual temperature dependent. So you can also take in considerations if it is a fair approximation to assume the density remains constant when you increase the temperature with 1350 degrees C

Answer 13

I was also given a melting point of: 1370 C
but no volume

Answer 14

Any chance you could take a pic of the question if it is from a book and attach it as a file? ect ect?

Answer 15

It seems like something is completely wrong, or I am just not aware of a possible solution.

Answer 16

I'm doing the extra credit. I don't really understand it though.

Answer 17

Yeah okay.... hmmm well first of the ones who wrote these questions this are not super genius. They constantly write the "heat capacity" but they constantly write what is known as the "specific heat capacity" which we define as:
\[\Large c _{s}=\frac{ C }{ m }\]
\(c_{s}\) is the specific heat capacity, \(C\) is the heat capacity, \(m\) is the mass.
From this definition, \(c_{s}\) gets the units joules per kelvin per gram (\(J K^{-1}g^{-1}\))

Answer 18

So it seems to me you are lacking information to solve the problem and they haven't throughout the question.

Answer 19

Yeah, sorry. My teacher kind of sucks at teaching chemistry obviously. I'm just giving up on the question.

Answer 20

Don't be sorry @jellyfishdream I just want to give you some good arguments if you confront them with the assignment, then they might also get a chance to correct it.

Answer 21

Thanks for everything @Frostbite