You are Here: Home >< Physics

# Uncertainties in specific heat capacity experiment. watch

1. I know that one example of uncertainty in determining the specific heat capacity of a liquid is that the temperature may not be uniform throughout the liquid, so stirring the liquid will reduce the uncertainty.

However the mark scheme also says that heat losses is another uncertainty. Apparently by using an initial temperature below room temperature and using the final temperature by the same amount above room temperature, helps to reduce uncertainty?
Can anyone explain the reasoning behind this one? it doesn't really make much sense to me.

Also another way of uncertainty is that some energy is required to raise temperature of the container / heater (etc), so by including the energy that is used to raise the container etc it in the calculation, you reduce the uncertainty? This is another one that I dont really understand?
2. (Original post by Fross877)
I know that one example of uncertainty in determining the specific heat capacity of a liquid is that the temperature may not be uniform throughout the liquid, so stirring the liquid will reduce the uncertainty.

However the mark scheme also says that heat losses is another uncertainty. Apparently by using an initial temperature below room temperature and using the final temperature by the same amount above room temperature, helps to reduce uncertainty?
Can anyone explain the reasoning behind this one? it doesn't really make much sense to me. ...
In regard to your question(s), the answer lies at the heart of the concept of balancing energy gains/losses before and after the experiment. The liquid is cooled to say about 5 °C below room temperature and then warmed it to about 5 °C above room temperature as the final temperature for the temperature can help to reduce the error because
- when the liquid is 5 °C below room temperature and the liquid is heated to room temperature, there is a certain of amount of heat gains by the liquid from the room, say +Q.
- when the liquid is at room temperature and continue to be heated to a temperature of 5 °C above room temperature, there is a certain of amount of heat losses from the liquid to the room. The amount of heat losses will be -Q.
Therefore, the heat gained by the liquid is balanced by the heat lost by liquid during the experiment.

(Original post by Fross877)
...
Also another way of uncertainty is that some energy is required to raise temperature of the container / heater (etc), so by including the energy that is used to raise the container etc it in the calculation, you reduce the uncertainty? This is another one that I dont really understand?
If the container is heated to the final temperature can reduce the error is again because of the balancing of heat lost and heat gain during the experiment. Try to explain yourself (very similar to the above explanation). If you can’t, let me know and I would post the explanation.
3. (Original post by Eimmanuel)
In regard to your question(s), the answer lies at the heart of the concept of balancing energy gains/losses before and after the experiment. The liquid is cooled to say about 5 °C below room temperature and then warmed it to about 5 °C above room temperature as the final temperature for the temperature can help to reduce the error because
- when the liquid is 5 °C below room temperature and the liquid is heated to room temperature, there is a certain of amount of heat gains by the liquid from the room, say +Q.
- when the liquid is at room temperature and continue to be heated to a temperature of 5 °C above room temperature, there is a certain of amount of heat losses from the liquid to the room. The amount of heat losses will be -Q.
Therefore, the heat gained by the liquid is balanced by the heat lost by liquid during the experiment.

If the container is heated to the final temperature can reduce the error is again because of the balancing of heat lost and heat gain during the experiment. Try to explain yourself (very similar to the above explanation). If you can’t, let me know and I would post the explanation.
For the 1st question I understand it now, however im still unsure about the second question.
4. (Original post by Fross877)
For the 1st question I understand it now, however im still unsure about the second question.
You may want to be more specific about what you are unsure. As I mention it is about balancing of heat lost by the container and heat gained by the container during the experiment.
5. (Original post by Eimmanuel)
You may want to be more specific about what you are unsure. As I mention it is about balancing of heat lost by the container and heat gained by the container during the experiment.
I was assuming that because it was separate points on the markscheme, that the previous point about having the initial temperature below room temp would not apply to the separate point about including the energy used to heat up the container. So I was a bit confused about the heat losses.
6. (Original post by Fross877)
I was assuming that because it was separate points on the markscheme, that the previous point about having the initial temperature below room temp would not apply to the separate point about including the energy used to heat up the container. So I was a bit confused about the heat losses.
It is indeed a separate point but the concept is the same.

If the container is heated to the final temperature say Tf (assume that the container does not lose thermal energy to the surrounding), the liquid (say at Ti) inside the container starts to be heated. The container will lose heat to the liquid because the liquid is at a lower temperature. At some point of time, the container and liquid will reach an equilibrium temperature Teq < Tf. When the liquid is continued to be heated, thermal energy will also be gained by the container. Remember previously, the liquid gains some heat from the container but now the liquid is “returning” back the heat it gains.
7. (Original post by Eimmanuel)
It is indeed a separate point but the concept is the same.

If the container is heated to the final temperature say Tf (assume that the container does not lose thermal energy to the surrounding), the liquid (say at Ti) inside the container starts to be heated. The container will lose heat to the liquid because the liquid is at a lower temperature. At some point of time, the container and liquid will reach an equilibrium temperature Teq < Tf. When the liquid is continued to be heated, thermal energy will also be gained by the container. Remember previously, the liquid gains some heat from the container but now the liquid is “returning” back the heat it gains.
Thank you that makes a lot more sense now

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: May 31, 2018
Today on TSR

Uni realities

### University open days

• University of Lincoln
Mini Open Day at the Brayford Campus Undergraduate
Wed, 19 Dec '18
• University of East Anglia
Fri, 4 Jan '19
• Bournemouth University