Finding the specific heat capacity of water using the gradient of a graph


by sungholee
Tags: calorimetry, lab report, specific heat
sungholee
sungholee is offline
#1
Nov6-12, 10:25 AM
P: 10
1. The problem statement, all variables and given/known data

Hi.
I have an assignment to find the specific heat capacity of water. We did an experiment in class where we used an electric kettle with power output of 1850W-2200W to heat up 1,400g of water (we actually used 1,400 mL of tap water but we were told to assume that the tap water has the same density as pure water). We used a temperature sensor and took recording every 10 seconds. We heated up the water for 240 seconds using the kettle.

Using the data that we got, we were supposed to draw a graph (what graph to use wasn't specified) and from the gradient of the graph, we had to find the specific heat capacity of water.


2. Relevant equations

Obviously, I used the equation Q=mcΔT and rearranged it to get c=Q/mΔT, where T is the temperature. Since we were using an electric kettle, Q=Pt where P is the power and t is the time. So, c=PΔt/mΔT


3. The attempt at a solution

I first tried graphing a graph with PΔt on the x-axis and mΔT on the y-axis. I did this by first using excel to get the Pt and mT rows and then graphing it. However, the gradient of this graph does not give me 4.18, but 0.2096.
So then, I tried graphing with only Δt on the x-axis and ΔT on the y-axis. Then, I multiplied the gradient (0.3031) by P/m, which was 2025 (the mean of the maximum and minimum power output of the kettle) divided by m, 1400g. However, I got 0.4384125 this time. Because some error from 4.18 was to be expected, I though that maybe I mixed up the units by a power of 10 somewhere since if 0.4384125 x 10 = 4.38... which is close enough. But P is in watts (J/s), t is in seconds, m is in grams and ΔT is in C (which should be different from using K since its the difference).

So I don't know where I went wrong. If I don't use the graph but just calculate using the averages ie. Pt/mT=(2025 x 240)/(1400 x 69.16), I get 5.019416674, which is still far from the real value but there were many sources of error in the design of the experiment.

In the end, my question is: Where am I going wrong? and how can I fix my graph?

I might not have been clear in some parts of my question because I'm not very familiar with forums so I'll be more than happy to clarify anything if you guys don't understand my post.

Thank you everyone.
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Chestermiller
Chestermiller is offline
#2
Nov6-12, 11:28 AM
Sci Advisor
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Thanks
PF Gold
P: 4,447
You must have made a mistake in the calculations somewhere. The only result that looks close is the one using the averages. The other two calculations are not consistent with this, or with eachother.

Here's a hint: Just plot T vs t, and see what it looks like. Is it a straight line relationship, or does it have a constant slope at long times? What is the slope of the straight line portion?

Can you provide a set of data so that we can see what they look like (not over the entire 240 seconds, but as T vs t, with 10 sec intervals on t)?
sungholee
sungholee is offline
#3
Nov6-12, 12:03 PM
P: 10
Hi Chestermiller. Thanks for your reply.

Here's my graph of T vs t.
Click image for larger version

Name:	T vs t.jpg
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As the graph shows, the gradient is 0.3031
I have the equation Pt/mT but if I multiply this by P which is 2025 and then divide by m which is 1400, I get 0.438... which isn't right.

nasu
nasu is offline
#4
Nov6-12, 12:49 PM
P: 1,904

Finding the specific heat capacity of water using the gradient of a graph


Quote Quote by sungholee View Post


Obviously, I used the equation Q=mcΔT and rearranged it to get c=Q/mΔT, where T is the temperature. Since we were using an electric kettle, Q=Pt where P is the power and t is the time. So, c=PΔt/mΔT


3. The attempt at a solution

I first tried graphing a graph with PΔt on the x-axis and mΔT on the y-axis. I did this by first using excel to get the Pt and mT rows and then graphing it. However, the gradient of this graph does not give me 4.18, but 0.2096.
You have PΔt = c (mΔT) so if you want c to be the slope of the graph you need to plot PΔt versus mΔT. This means mΔT on the x axis.
What you got is 1/c.
Chestermiller
Chestermiller is offline
#5
Nov6-12, 01:27 PM
Sci Advisor
HW Helper
Thanks
PF Gold
P: 4,447
Check your equation. You need to divide by the slope (0.3031), not multiply.
sungholee
sungholee is offline
#6
Nov6-12, 01:49 PM
P: 10
Thank you both nasu and Chestermiller so much!
I've been making a stupid mistake :p
I though I checked it several times but I just realized that I mixed up the x and y wrong and t and the T wrong. I don't even know why since it's so simple. I feel stupid now :p
Again, thank you so much!


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