How to find the specific heat of a metal?

Question

anonymous · Answer

This is the work I have so far 

Calculate the energy change (q) of the surroundings (water). We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL.
(27.776g)x(4.18j/g)x(25.3) =2944kj
Calculate the specific heat of the metal.

aaronq · Answer

you can find the specific heat by using the same equation, given that you know all the other variables.

anonymous · Answer

@aaronq I think I did the first part wrong, can you help me check my work?

anonymous · Answer

So i re-did it and this is my work: 
(27.776)(4.18)(13.6)= ?? 

I'm confused on what my answer would be

aaronq · Answer

1579.0 J

you need a calculator for these questions

anonymous · Answer

I used a calculator and I got that, but it seemed wrong so thank you!

aaronq · Answer

no problem. So for the metal (assuming that a hot metal was immersed in the water) the heat lost from it is the same as that gained by the water.

anonymous · Answer

Okay thank you! I'm about to calculate the specific heat, will you check that also?

aaronq · Answer

sure thing

anonymous · Answer

Do you happen to know an equation for this? I cant figure out how to do it @aaronq

aaronq · Answer

it's the same equation

$q=m*C_p*\Delta T$

you have to know the mass of the metal and the change in temp

anonymous · Answer

What does the Cp and the ΔT mean?

anonymous · Answer

*what do

aaronq · Answer

$C_p$ is the specific heat capacity 
$\Delta T$ is the change in temp

anonymous · Answer

Okay so the mass is 27.776 and the change in temp is 13.6

anonymous · Answer

Would the specific heat capacity be 4.18?

aaronq · Answer

27.776 g is also the mass of the metal?

anonymous · Answer

Yes here is my data table

anonymous · Answer

I'm sorry the temp of the metal is actually 100.5 degrees

anonymous · Answer

@aaronq

aaronq · Answer

Okay, so the first answer is wrong. 

You first have to find the heat exchanged through the change in temperature of the water.

aaronq · Answer

Use the mass of the water, it's change in temp and it's heat capacity

anonymous · Answer

Do I multiply all those together?

aaronq · Answer

yeah, because $q=m*C_p*\Delta T$

anonymous · Answer

Okay and just to clarify the heat capacity is 4.18? And whats the mass of the water?

aaronq · Answer

you have 26 mL and it's density is 1 g/mL

$\rho=\dfrac{m}{V}\rightarrow m=\rho*V=(1 ~g/mL)*(26~mL)=26~g$

anonymous · Answer

Okay so its 1478.0?

anonymous · Answer

@aaronq

aaronq · Answer

Did they tell you what's happening? was the hot Al dropped into a beaker?

anonymous · Answer

I'll check

aaronq · Answer

k

anonymous · Answer

No I don't think it wasn't, but once it was in the beaker it became hot because I heated it up

anonymous · Answer

*It was

aaronq · Answer

so when they say "the temperature of the beaker", do they mean the water?

anonymous · Answer

yes

aaronq · Answer

So the water was initially at 100 celsius and the metal at 25.3 celsius?

anonymous · Answer

Yes

anonymous · Answer

I think i figured it out, can you check it?

aaronq · Answer

okay, post it

anonymous · Answer

4.18/(24)(75.1) =13.07

aaronq · Answer

hm 24?

aaronq · Answer

I can't really tell what you're doing

anonymous · Answer

I used the average volume

anonymous · Answer

which was 24

anonymous · Answer

and I multiplyed that by 1.00 g/mL

aaronq · Answer

oh okay. I'm just saying because that's not what it says on the data you posted.

are you finding q above?

anonymous · Answer

Yeah, it would be 13.07 right?

aaronq · Answer

nope. you divided 4.18 for some reason.

You need to multiply them through.

q=(4.18)(24)(75.1)=7534.032 J

aaronq · Answer

now that you know q, find $C_p$ of the metal.

anonymous · Answer

Okay so I emailed my teacher and she gave me the correct equation which is, q(water) = 26 x 13.6 x 4.18. I must have gotten confused with the one above

aaronq · Answer

okay, so the metal was at 100 celsius, not the water.

okay, so now find the heat capacity of the metal

anonymous · Answer

1478 j?

aaronq · Answer

yep

anonymous · Answer

Okay so how do I find the specific heat?

aaronq · Answer

use the same equation with the variables of the metal

anonymous · Answer

1478 / (-61.6 x 27.776 ) ?

anonymous · Answer

*-1478

anonymous · Answer

I got 0.864 j

aaronq · Answer

you need to switch signs because the loss of heat by the metal is exothermic.

-1478 J = (27.776 g)*$C_p$*(-61.6)

aaronq · Answer

yeah that should be right. the units are not though

anonymous · Answer

What do the units need to be?

aaronq · Answer

J/g*K  (or celsius)

anonymous · Answer

oh okay thank you so much

aaronq · Answer

no problem !