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

In summary, the conversation discusses the process of finding the specific heat capacity of water using an experiment with an electric kettle. The equation Q=mcΔT is used and rearranged to get c=Q/mΔT. There were some errors in the calculations and graphing, leading to a discussion on the correct equation and plotting the correct variables. Eventually, the mistake was discovered and the correct values were obtained.
  • #1
sungholee
19
0

Homework Statement



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.

Homework 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

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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  • #2
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 each other.

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)?
 
  • #3
Hi Chestermiller. Thanks for your reply.

Here's my graph of T vs t.
T vs t.jpg


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.
 
  • #4
sungholee said:
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


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.
 
  • #5
Check your equation. You need to divide by the slope (0.3031), not multiply.
 
  • #6
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!
 

1. What is the specific heat capacity of water?

The specific heat capacity of water is the amount of heat energy required to raise the temperature of 1 gram of water by 1 degree Celsius.

2. How is the specific heat capacity of water measured?

The specific heat capacity of water is typically measured using the gradient of a graph. This involves plotting the change in temperature of water against the amount of heat energy added to it, and calculating the slope of the resulting line.

3. Why is the gradient of a graph used to find the specific heat capacity of water?

The gradient of a graph is used because it represents the relationship between the change in temperature and the amount of heat energy added to the water. By calculating the gradient, we can determine the specific heat capacity of water.

4. What is the unit of measurement for specific heat capacity?

The unit of measurement for specific heat capacity is Joules per gram per degree Celsius (J/g°C).

5. Can the specific heat capacity of water vary?

Yes, the specific heat capacity of water can vary slightly depending on factors such as temperature and impurities in the water. However, it is generally accepted to be around 4.18 J/g°C.

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