Question

Help with a chemistry assignment?

Answer 1

Part I:

Calculate the energy change (q) of the surroundings (water) using the enthalpy equation

q\(_{water}\) = m × c × ΔT.

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.

The water has absorbed the heat of the metal. So, q\(_{water}\) = q\(_{metal}\)

Using the formula qmetal = m × c × ΔT, calculate the specific heat of the metal. Use the data from your experiment for the metal in your calculation.
Part II:

Calculate the energy change (q) of the surroundings (water) using the enthalpy equation

q\(_{water}\) = m × c × ΔT.

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.

The water has absorbed the heat of the metal. So, q\(_{water}\) = q\(_{unknown}\) metal

Using the formula q\(_{unknown ~metal}\) = m × c × ΔT, calculate the specific heat of the metal. Use the data from your experiment for the unknown metal in your calculation.

Answer 2

It's actually, \(\sf \Large q_{water}=-q_{metal}\), if the metal is hotter than the water

Answer 3

Here are my notes from the virtual lab:

Copper
Measured Mass of Metal: 41.664 g
Distilled Water Measurement: 26 mL
Distilled Water Temperature: 25.4 Cº
Temperature of Metal: 100.7 Cº
Temperature of Mixture: 34.5 Cº

Aluminum
Measured Mass of Metal: 41.664g
Distilled Water Measurement: 26 mL
Distilled Water Temperature: 25.4 Cº
Temperature of Metal: 100.7 Cº
Temperature of Mixture: 44.1 Cº

Zinc
Measured Mass of Metal: 41.664g
Distilled Water Measurement: 26mL
Distilled Water Temperature: 25.4 Cº
Temperature of Metal: 100.7 Cº
Temperature of Mixture: 34.8 Cº

Iron
Measured Mass of Metal: 41.664g
Distilled Water Measurement: 26 mL
Distilled Water Temperature: 25.4 Cº
Temperature of Metal: 100.7 Cº
Temperature of Mixture: 25.7 Cº

Unknown Metal 1
Measured Mass of Metal: 25.605 g
Distilled Water Measurement: 24.8 mL
Distilled Water Temperature: 25.2 Cº
Temperature of Metal: 100.5 Cº
Temperature of Mixture: 0 Cº

Unknown Metal 2
Measured Mass of Metal: 25.605g
Distilled Water Measurement: 24.8 mL
Distilled Water Temperature: 25.2 Cº
Temperature of Metal: 100.5 Cº
Temperature of Mixture: 32.2 Cº

Unknown Metal 3
Measured Mass of Metal: 25.605 g
Distilled Water Measurement: 25.2mL
Distilled Water Temperature: 24.8 Cº
Temperature of Metal: 100.5 Cº
Temperature of Mixture: 28.7 Cº

Answer 4

I have all of the measurements, it's just I'm not sure what to use in the formula

Answer 5

The formula is given in the question, \(\sf q_{water} = m × c × ΔT\)

where \(q_{water}\) is the heat absorbed (in this question) by the water
m is mass (of water, here)
c is specific heat capacity
\(\Delta T\) is the change in temperature, that is \(\Delta T=T_f-T_i\)

Answer 6

Am I only supposed to use one of the known metals and one unknown, or all of them do you think?

Answer 7

To find the heat capacity of the metal, you'd use the same formula, but different temperatures, and the heat quantity would be negative because heat was released.

Answer 8

I think they want you to do it for every set of data

Answer 9

When you do the unknown metals, you'd find \(q_{water}\), then you'd know \(q_{metal}\) (it's the same, except the sign change).

You'd then use \(q_{metal}\) to find the specific heat capacity of the metal.

Answer 10

So i'll show you an example with copper

Copper
Measured Mass of Metal: 41.664 g
Distilled Water Measurement: 26 mL
Distilled Water Temperature: 25.4 Cº
Temperature of Metal: 100.7 Cº
Temperature of Mixture: 34.5 Cº

\(q_{water}=m*c*\Delta T=26 ~g*4.18~J/gC^o*(34.5C^o-25.5C^o)=978.12~J\)

Answer 11

then \(q_{metal}=-978.12~J=41.664~g*c*(34.5C^o-100.7C^o)\)

c=0.35 ~J/g C^o

Answer 12

Okay, so using aluminum

Aluminum
Measured Mass of Metal: 41.664g
Distilled Water Measurement: 26 mL
Distilled Water Temperature: 25.4 Cº
Temperature of Metal: 100.7 Cº
Temperature of Mixture: 44.1 Cº

\(\sf q_{water} = 26 × 4.18 J/gCº × (44.1 Cº−25.4 Cº)\)

c=2032.316 J/g Cº

Answer 13

There's another part too if you'd be willing to help me with that?

Answer 14

You actually found \(q_{water}\), not c

c is found in an additional step where you use the data for the metal

Answer 15

oh, would this be the additional step?
\(\color{blue}{\text{Originally Posted by}}\) @aaronq
then \(q_{metal}=-978.12~J=41.664~g*c*(34.5C^o-100.7C^o)\)

c=0.35 ~J/g C^o
\(\color{blue}{\text{End of Quote}}\)

Answer 16

yeah

Answer 17

why -978.12? wouldn't it be positive?

Answer 18

because the heat is transferred, meaning one object released it and another absorbed it.

Because of this directionality, there is convention, heat absorbed is positive and released negative.

thus in this case, \(\sf q_{water}=-q_{metal}\)

Answer 19

I'm really sorry, this is just confusing for me

how do you get from -978.12 to 41.664gc?

Answer 20

so in that "additional step", i used the formula again - with different data.

\(q_{metal}=m*c*\Delta T\)

\(−978.12 J=(41.664~ g)∗c∗(34.5Co−100.7Co)\)

Answer 21

ooohhhh that makes sense now

Answer 22

okay coool. let me know if you still need help

Answer 23

Could you possibly help me on the last part? I can open a new post if that would be easier