Question

Part II:
1. 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.
2. Calculate the specific heat of the metal.
Conclusion:
1. Use the given specific heat capacity values below to calculate the percent error of the experimental specific heat capacity that you determined in Part I of the lab.
Known specific heat values—Iron: 0.444 J/g°C; Zinc: 0.390 J/g°C; Copper: 0.385 J/g°C, Aluminum: 0.900 J/g°C

Answer 1

1. (27.776 g) * ( 4.18 J/g *C ) * ( 25.3 C )
Q=2944kJ

Answer 2

Q = s(m)(delta T)
Q = specific heat of X (mass of X) (temp difference)
Since we don't know the whole question, ie starting and ending T, we cannot tell you delta T.
We cannot check #1 for the same reason.

However: specific heat of water is usually 4.186 j/g C, and density is 1g/ml, so it looks right.
Look up the specific heat of whatever metal you are using. Since we don't know the T range, we just have to assume your value of 25.3 is correct. However, the metal may have a different boiling point from water, you should look that up. Since we don't know your starting point you'll have to figure that out too. If room temp, 25C is considered room temp unless otherwise specified.

Answer 3

I'll try get your other answers

Answer 4

oh my god you are a lifesaver!!!!!!!!!!!!!!!!!!!!!!!

Answer 5

haha :)

Answer 6

Here I'll show you the work I have

Answer 7

and this is what I need: Use the given specific heat capacity values below to calculate the percent error of the experimental specific heat capacity that you determined in Part I of the lab.
Known specific heat values—Iron: 0.444 J/g°C; Zinc: 0.390 J/g°C; Copper: 0.385 J/g°C, Aluminum: 0.900 J/g°C

Using the experimental specific heat capacity value that you determined in Part II of the lab, what is the most probable identity of the metal that you examined?

Assuming that is the identity of the metal, determine the percent error of your calculated specific heat capacity value.

Answer 8

This is the chart I use for reference:

Answer 9

These are my charts

Answer 10

Known specific heat values — Iron: 0.444 J/g°C; Zinc: 0.390 J/g°C; Copper: 0.385 J/g°C, Aluminum: 0.900 J/g°C

| experimental - actual value | x 100 %
actual value

-
experimental value = 0.39
actual value (known) = 0.39

Now you plug in.

9(.39-.39)x100%)/.39 = 0%

So, you have 0% error. That would make sense, given that your experimental results was THE SAME as the known value. It's the same. There's no difference.

There is no error. You have 0% error. this part is for them one you just sent right now

Answer 11

sorry sorry those are your charts?

Answer 12

yes :)

Answer 13

The written ones are, the other chart was part of the lesson

Answer 14

9(.39-.39)x100%)/.39 = 0%

I don't have much time to help you but I can come back later and help but If you use that equation and substitute your values in your charts you will get your answers for your question

Answer 15

I'll try my best right now to give as much info to help you as I can whats the thing your are most unsure with?

Answer 16

It's so very much apprciated, really

Answer 17

I'm confused though, how did you get .39?

Answer 18

.39 is the actual percentage I got that before you gave me your chart thats an example so you would plug your chart work in that equation

Answer 19

oh, that was a mental question to ask, I'm sorry. Also, did you say I had similar results to what was the actual value of the experiment?

Answer 20

| experimental - actual value | x 100 %
actual value

Answer 21

yes you do

Answer 22

Okay, your time has been very appreciated! you're fabulous!!!!!!

Answer 23

I'm so sorry my work is kinda confusing but if you read the words it explains a bit how to do it I'll come back later to help out if you need anything else bye bye thank you

Answer 24

No, I don't have a single complaint for your work; it helped me very very much!

Answer 25

thank you :) did you get this assignment?