A 410g cylinder of brass is heated to 95.0°C and placed in a calorimeter containing 335g of water at 25.0°C. The water is stirred and its highest temperature is recorded at 32.0°C. From the thermal energy gained by the water, determine the specific heat of brass. The specific heat of water is 4.18 J/g · °C

Question

A 410g cylinder of brass is heated to 95.0°C and placed in a calorimeter containing 335g of water at 25.0°C. The water is stirred and its highest temperature is recorded at 32.0°C. From the thermal energy gained by the water, determine the specific heat of brass.  The specific heat of water is 4.18 J/g · °C

leonardo0430 · Answer

what would be the question?

haleyelizabeth2017 · Answer

I have to find the specific heat of brass....I just need help getting started then someone to check my answer after I finish...

matt101 · Answer

Whatever heat is lost by the brass is gained by the water. That means:

$$m_{brass} c_{brass} \Delta T_{brass}=m_{water} c_{water} \Delta T_{water}$$

haleyelizabeth2017 · Answer

oh

haleyelizabeth2017 · Answer

mass is m and temperature? is t, correct?

leonardo0430 · Answer

basically what matt said use the equation to determine the end product

leonardo0430 · Answer

correct

haleyelizabeth2017 · Answer

So I just plug everything and solve for c

haleyelizabeth2017 · Answer

Okay....give me a second....would you mind checking my answer once I get it?

leonardo0430 · Answer

yes so just plug in the equation, sure ill check it

haleyelizabeth2017 · Answer

Okay, thank you

haleyelizabeth2017 · Answer

Do i do the final temperature or beginning temperature or temp difference?

leonardo0430 · Answer

the beginning

haleyelizabeth2017 · Answer

Okay....

matt101 · Answer

∆T is the temperature difference!

Final temperature minus initial temperature

haleyelizabeth2017 · Answer

Okay.

haleyelizabeth2017 · Answer

One second while I make a few adjustments ;)

haleyelizabeth2017 · Answer

Do I do the same to the brass because we know that it lost 7 degrees....since it transferred to the water

leonardo0430 · Answer

yes do the same :)

haleyelizabeth2017 · Answer

So what I did was:
(410)(88)c=(335)(4.18)(7)
36,080c=9,802.1
c=0.27

Did I do it right?

matt101 · Answer

NO @leonardo0430 !

Just because the water GAINED 7 degrees, doesn't mean the brass LOST 7 degrees. This would imply that specific heats of both substances are the same, which is not the case.

Eventually the water and brass reach thermal equilibrium when they are BOTH the SAME TEMPERATURE. If the highest (i.e. final) temperature of the water is 32 degrees, then that is also the final temperature of the brass! If the brass were still 88 degrees when the water is 32 degrees, thermal equilibrium has not been reached. The brass would continue transferring heat to the water, and the temperature of the water would continue to rise.

haleyelizabeth2017 · Answer

Ugh this is so confusing :(

haleyelizabeth2017 · Answer

What do I do then @matt101?

haleyelizabeth2017 · Answer

Ohhh! 410*63?

haleyelizabeth2017 · Answer

Then I would get about 0.380....

matt101 · Answer

The CHANGE in temperature is what we want.

For water: final temp = 32, initial temp = 25, change in temp = 32-25 = 7
For brass: final temp = 32, initial temp = 95, change in temp = 32-95 = -63

matt101 · Answer

Yes!

haleyelizabeth2017 · Answer

okay.....

haleyelizabeth2017 · Answer

Thank you

matt101 · Answer

Btw, I should add to the equation I wrote out at the beginning that there should be a - on one side, because the heat gained by one side is lost by the other (on side is a positive change the other is a negative change).

Having this negative in the equation will cancel the negative from the change in temperature so that your c will be positive.

And yes, c=0.38!

haleyelizabeth2017 · Answer

Okay..... :)