Question

in a calometry experiment, .50 kg of a metal @ 100 degrees Celsius is added to .50 L of water @ room temperature in an aluminum calometer cup. The cup has a mass of 250 g. a) if the final temperature of the mixture is 25 degrees Celsius, what is then specific heat of the metal? b) What would be the effect if some water splashed out of the cup when the metal was added?

Answer 1

You'll need the specific heat of water and of aluminum to answer this question, so look those up if you don't have them.

Answer 2

specific of water is 1 and aluminum 0.21

Answer 3

*specific heat rather

Answer 4

To answer this, we use the formula:
\[Q = mC\Delta T\]
If we have multiple sources for heat transfer:
\[Q = (m_1C_1+m_2C_2+...)\Delta T\]

Since heat is energy, and energy is conserved (we're assuming heat doesn't actually leave the calorimeter), then the heat leaving the metal equals the heat that enters the water and the aluminum calorimeter.
\[m_{metal}C_{metal}\Delta T_{metal} = (m_{water}C_{water} + m_{aluminum}C_{aluminum})\Delta T_{water+calorimeter}\]
Now, "delta T" is the change in temperature:
\[\Delta T = T_{final}-T_{initial}\]
Substitute for that, plug in your initial temperatures, and solve for the final temperature.

Answer 5

Oops, sorry. Just find delta T, and solve for C metal. :)

Answer 6

in ΔT water + calorimeter , what will substitute in that calorimeter?

Answer 7

uhm,,,what's the specific heat for metal? i can't find it ..

Answer 8

\[\Delta T_{metal} = 100 - 25 = 75\]
\[(m_{water}C_{water}+m_{alum}C_{alum})(T_f-T_i) = m_{metal}C_{metal}(75)\]
The water and cup were initially at room temperature. T final will be 25 as well (When in equilibrium, everything is the same temperature). Plug those numbers in, solve for C metal.

Answer 9

okay,, :)

Answer 10

I didn't do delta T for the water and cup, because "room temperature" definitions vary :)

Answer 11

if i use 20 degrees celsius?

Answer 12

.50L is 500 grams? right?

Answer 13

That's reasonable, I suppose. It's usually mentioned in class at some point.

Yeah, .5 L = 500 g for water

Answer 14

what would be my initial temp.?100 degrees celsius?

Answer 15

For the water and calorimeter, your initial temperature is room temperature. It tells you that the final temperature is 25 degrees.

Answer 16

(500 g (1) + 250g (0.21)(25-20) = 500gg Cmetal (75) ...right?

Answer 17

Looks good to me (you missed a parenthesis on the left, though)

Answer 18

in 500g?

Answer 19

Just after the (0.21), before the (25-20)

Answer 20

What would be the effect if some water splashed out of the cup when the metal was added?

Answer 21

what will be my unit in my answer? :)

Answer 22

Well, if water splashes out, then we have less water absorbing the heat from the metal. The final temperature would then be higher than 25, as the remaining water will have to absorb some more energy to reach thermal equilibrium with the metal.

Answer 23

The units should be that of a specific heat: cal/g K

Answer 24

ahhhh,,,thank youu :)

Answer 25

ahhh ...

Answer 26

after this,,,


2762.5 = 500g Cmetal (75), whats next?

Answer 27

transpose 500g on the left?

Answer 28

Oh, I gave the units for your answer for C. What you have on the left hand side has units of calories (energy).

Now, you just solve for C: 500*75*C = 2762.5
\[C = \frac{2762.5}{500*75}\]

Answer 29

ahhhh,,,gets !!