Question

Calorimetry help please! I need a walkthrough on how to do this two-part question.

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

qwater = 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, qwater = qmetal
(I'll post second part in a comment.)

Answer 1

2. 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.

The data on my element is mass: 27.776, Volume of water: 26 mL, Water temperature: 25.3*C, Metal and water: 100.5*C, Final temperature: 38.9*C
I have to compare the results of two elements. The second element is: mass: 15.262g, Volume of water: 24.0 mL, Temperature of water: 25.2*C, Metal and water temperature: 100.3*C, Final 27.5*C

Please help me because I have been stuck on this assignment for two weeks. I just do not get it.

Answer 2

Is this two questions?

Answer 3

Like I'm a bit confused on what your asking.

are your two questions based on

"The data on my element is mass: 27.776, Volume of water: 26 mL, Water temperature: 25.3*C, Metal and water: 100.5*C, Final temperature: 38.9*C
I have to compare the results of two elements. The second element is: mass: 15.262g, Volume of water: 24.0 mL, Temperature of water: 25.2*C, Metal and water temperature: 100.3*C, Final 27.5*C"?

Answer 4

Yes, the two questions are based on that information. Like she wants me to plug in the data of those two elements into each question. So it's basically a four-part question.

Answer 5

So each question, solve for each element?

Answer 6

Exactly. I'm sorry, I know it's a lot :(

Answer 7

Hmm, I'll do one element and solve for both questions, and you do the other element, okay?

Answer 8

Sounds like a plan :D

Answer 9

The first element: "The data on my element is mass: 27.776, Volume of water: 26 mL, Water temperature: 25.3*C, Metal and water: 100.5*C, Final temperature: 38.9*C
I have to compare the results of two elements. "

Solving for the energy change of the water..

you need the change in temperature, specific heat, and mass.

mass=26 grams as the density is 1.

d=m/26

m=26 grams

specific heat is given...at 4.18 j.

change in temperature is 38.9-25.3=13.6

13.6 x 4.18 x 26=1478 joules

2nd question for the same element

"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."

"The data on my element is mass: 27.776, Volume of water: 26 mL, Water temperature: 25.3*C, Metal and water: 100.5*C, Final temperature: 38.9*C"

solving for C (specific heat)

m=27.776
change in temperature(with metal)=38.9-100.5=-61.6

using the 1478 joules from the first question...

1478=(-61.6)(27.776)(specific heat)

specific heat=roughly 0.864 J

Answer 10

take your time and read that over, and solve for the 2nd element. it's not that hard on paper, I just wrote everything out (even the unecessary stuff)

Answer 11

1. M=26g
25.3-27.5=2.2
2.2x4.18x26=239.09J

2. M=15.262g
27.5-100.3=-72.8
239.09=(-72.8)(15.262g)

So I have everything up to there. Do you mind me asking what you did in order to get your specific heat? I'm horrible with equations xD

Answer 12

Thank you so much btw. This is the first time any of this jibberish has made sense to me.

Answer 13

1. M=26g
25.3-27.5=2.2
2.2x4.18x26=239.09J

2. M=15.262g
27.5-100.3=-72.8
239.09=(-72.8)(15.262g)

So I have everything up to there. Do you mind me asking what you did in order to get your specific heat? I'm horrible with equations xD

Answer 14

1. M=26g
25.3-27.5=2.2
2.2x4.18x26=239.09J

2. M=15.262g
27.5-100.3=-72.8
239.09=(-72.8)(15.262g)

So I have everything up to there. Do you mind me asking what you did in order to get your specific heat? I'm horrible with equations xD

Answer 15

Sorry, I didn't meant to post that a million times. My cat sat down on my laptop when I walked away for a second, haha.

Answer 16

"The second element is: mass: 15.262g, Volume of water: 24.0 mL, Temperature of water: 25.2*C, Metal and water temperature: 100.3*C, Final 27.5*C"

the mass would be 24 grams.

you put 26 for some reason, otherwise that is right

Answer 17

for the specific heat of the metal, first you have to fix the first question

you did most of that right when solving for specific heat, it's basically just like algebra here.

(just an example below)

2352=(231)(3)(specific heat)

231x3=693

2352=693(specific heat)

divide both sides by 693

specific heat=3.39

Answer 18

Okay, I think I've got it. Let me re do the question and I'll post it here in a second.

Answer 19

1. M=24g
2.2x4.18x24=220.704

2. 220.709=(-72.8)(15.262g)(specific heat)
220.704=-1,111.0736(specific heat)
SH=-5.03422

Answer 20

i'll check this in 30 min, gotta run sorry

Answer 21

the problem with your 2nd question is that when you have

220.704=1,111.736(specific heat)

you have to divide both sides by 1,111.736 so you can isolate the specific heat

specific heat=0.1987

Answer 22

Ohhh, I see. I divided it by the other number. Thank you for all your help. You are awesome and I understand how to work these now :)

Answer 23

Yw, feel free to fan and medal. Tag me onto more questions too.

Answer 24

I fanned, and I made you best answer. I think that is a medal? xD