# Specific Heat Practice

This is practice for a test but I'm confused so I wanted to ask a few questions before taking it. A total of 74.8 J of heat is required to raise the temperature of 18.69 g of silver from 10.0 to 27.0 oC.

What is the specific heat of silver? Show the equation, your substitution, and the answer.

Do I use this equation? Heat = mass * specific heat * change in temp (Using words instead of symbols/letters is easier on here.)

If so, how exactly do I do it? I tried it but when I checked my answer I was way off. The answer the teacher has down is "74.8 g = .235 J/goC"

## Answers and Replies

Pengwuino
Gold Member
Well you basically have this equation

$$E = mc\Delta T$$

m = mass
c = specific heat
delta T = change in temperature

Since you are looking for specific heat, you re-write the equation as

$$c = \frac{E}{{m\Delta T}}$$

I don't understand what's going on with the left side of the equation you provided for the answer, 74.8 g can't be equal to something that has units of J/(g C). The right side of the equation is the correct answer for the specific heat though, and you do use the formula that you gave.

I dunno...I copied and pasted what he had down. ^_^

I knew I had to rearrange the equation but I couldn't seem to get it just right. Not sure why though, I'm usually pretty good with algebra...maybe I'm just tired.:rofl: Either that or I'm confused because the equations flip letters every once in a while...one minute it's q, the next it's e...one minute specific heat is s, the next it's c.:uhh:

Ah, I just figured out what I was doing wrong...I was trying to rearrange it all wacky like.:tongue:

Q = amount of heat
E = Energy (in this context, the heat energy = amount of heat)

s = Probably short for specific heat capacity in written text :S
c = The physical variable of specific heat capacity

GCT
Science Advisor
Homework Helper
show your work and we can go on from there, this problem is as simple as it gets....you plug in the numbers and check for the units, you may have gone wrong with the units somewhere.

Sorry it took so long for me to reply, trying to do this and balance Hamlet at the same time is a bit hard. I got it wrong again but I can't figure out what I'm doing wrong. I know it's simple but I still can't get it. I was given this equation: q = msT (Had to let out the triangle but you all get the gist of the equation.)

I changed it to: s = q/mT for this problem:

A total of 74.8 J of heat is required to raise the temperature of 18.69 g of silver from 10.0 to 27.0 oC.

What is the specific heat of silver?

s = 74.8/18.69(17) = 4.25

Answer? 74.8 g = .235 J/goC Here is the solution:

$$E = 74.8 J$$

$$m = 18.69 g = 18.69 * 10^-^3 kg$$

$$\Delta T = 27.0 - 10.0 = 17.0$$

$$c = ?$$
-----------------------------------------------------------
$$E = mc\Delta T$$

$$c = E/m\Delta T = 74.8/(18.69 * 10^-^3 * 17)$$

$$c = 235J/kg/K$$

Answer is rounded to 3 significant figures and AngelShare, always use SI-units.

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Where did the 10^-3kg come from?

converting from grams to kilograms? its part of the scientific notation

Ah, that was a stupid mistake I made. I wasn't paying attention and put 317.73/74.8 into my calculator instead of 74.8/317.73.

As for converting my answer from grams to kilograms, the answer is in grams so it's fine just the way it is. ^_^

Actually it is not AngelShare. It is bad practice not to use SI-units. You will see it in more advanced formulas.

On tests (for me atleast) points get taken for not using SI-units (in calculations and answer)

Well, he had the answer down exactly as I posted it here so he didn't convert it...so he won't be taking points away if we don't. Now that I think about it, I don't remember doing anything like that yet, I may have to read up about it in my books because I haven't encountered it yet in this course...or at least I don't think I have, I'm rather tired right now, I was just trying to get another assignment done before heading off to bed. ^_^