Question

The specific heat of a certain type of metal is 0.128 J/(g·°C). What is the final temperature if 305 J of heat is added to 86.1 g of this metal initially at 20.0 °C?

Answer 1

You know how to related the energy given to the heat capacity?

Answer 2

I do not know how to.
What is the equation to do the problem, that's what I cannot remember.

Answer 3

Right the equation we are going to be using is:
\[\Large Q=m \times c \times \Delta T\]
Where Q is the heat added, m is the mass, c is the specific heat capacity and \(\Delta T\) is the difference in temperature expressed as \(\Delta T=T_{2}-T_{1}\) .

Answer 4

And welcome to \(\LARGE \sf \bbox[#40B9E9]{\color{white}{Open}}\bbox[#A8CE91]{\color{white}{Study}}\) by the way.

Answer 5

Try solve for \(T_{2}\) first.

Answer 6

Let me help you a bit on the way:
\[\Large \Delta T=T _{2}-T _{1}=\frac{ Q }{ m \times c }\]

Answer 7

Okay, so I am solving for T2 first. T1=20.0, Q= 305 J, m+86.1, and 0.128 J/(g·°C)

Answer 8

Is that correct?

Answer 9

Oh, and thank you! :)

Answer 10

That is correct the values you've set. :)

Answer 11

Yay! So do I need to change 0.128 to just degrees celcius or is it okay as is?

Answer 12

It should be okay to hold it as celsius as all the units cancel out to celcius in the end.

Answer 13

But if you come with an answer I'll be more than happy to check it.

Answer 14

Sorry, my computer died. I got 27.67 for the answer.. and I do not think that that is correct.. :/

Answer 15

Hey, so did mine.
\[\large T _{2}=\frac{ 305 ~ \textrm{J} }{ 86.1 ~ \textrm{g} \times 0.128 \frac{ \textrm{J} }{ \textrm{g} \times ^{\circ} \textrm{C} } }+20.0 ^{\circ} \textrm{C}=27.67 ^{\circ} \textrm{C}+20.0 ^{\circ} \textrm{C}=47.67 ^{\circ} \textrm{C}\]
You just forgot to add \(T_{1}\) :)

Answer 16

I said I would only give a hit so the last equation I gave you should be rearranged to:
\[\Large T _{2}=\frac{ Q }{ m \times c }+T_{1}\]
But you did it correctly. you calculated how much the temperature would increase from it's current state.