GCSE Physics Specific Latent Heat Textbook Question Watch

astudentastudent
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Please help me understand this question and how to work it out.

An ice cube of mass 0.008kg at 0 degrees Celsius was placed in water at 15 degrees Celsius in an insulated plastic beaker.The mass of water in the beaker was 0.120kg.After the ice cube had melted,the water was stirred,and its temperature was found to have fallen to 9 degrees Celsius.The specific heat capacity of water is 4200J/Kg degrees Celsius.
Q:Calculate the energy transferred from the water-2 marks.
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Callicious
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(Original post by astudentastudent)
Please help me understand this question and how to work it out.

An ice cube of mass 0.008kg at 0 degrees Celsius was placed in water at 15 degrees Celsius in an insulated plastic beaker.The mass of water in the beaker was 0.120kg.After the ice cube had melted,the water was stirred,and its temperature was found to have fallen to 9 degrees Celsius.The specific heat capacity of water is 4200J/Kg degrees Celsius.
Q:Calculate the energy transferred from the water-2 marks.
The ice cube has a mass, and it has a specific latent heat capacity. We also know that it changes temperature, giving us a change in temperature of (final - initial).

Using the formula of E = mcΔT, we can find out the energy lost (transferred from the water); E = energy, m = mass, c = specific heat capacity, the last part = change in temperature. I hope I helped!
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Eimmanuel
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(Original post by Callicious)
The ice cube has a mass, and it has a specific latent heat capacity. We also know that it changes temperature, giving us a change in temperature of (final - initial).

Using the formula of E = mcΔT, we can find out the energy lost (transferred from the water); E = energy, m = mass, c = specific heat capacity, the last part = change in temperature. I hope I helped!
Do you realize that your value of the energy lost is negative, which differ in some way that the question is asking?
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Callicious
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(Original post by Eimmanuel)
Do you realize that your value of the energy lost is negative, which differ in some way that the question is asking?
Well, if it's negative then that is simply to show that energy has been taken out of the system at hand... best explanation I have. If that's a problem, just use the formula with a positive difference.
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Eimmanuel
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(Original post by Callicious)
Well, if it's negative then that is simply to show that energy has been taken out of the system at hand... best explanation I have.

I am not sure if you understand the significance of negative value here or you are just confused.

Based on the question, it is asking about energy transferred from the water which is equivalent to your energy lost in post 2.

If "energy lost from the system" is say -10 J, it means that the system is gaining 10 J of energy instead of losing 10 J of energy.

If "energy lost from the system" is say 50 J, it means that the system is losing 50 J of energy and the energy gained from the system is -50J.

Sign is important! Be careful.
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Callicious
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(Original post by Eimmanuel)
I am not sure if you understand the significance of negative value here or you are just confused.

Based on the question, it is asking about energy transferred from the water which is equivalent to your energy lost in post 2.

If "energy lost from the system" is say -10 J, it means that the system is gaining 10 J of energy instead of losing 10 J of energy.

If "energy lost from the system" is say 50 J, it means that the system is losing 50 J of energy and the energy gained from the system is -50J.

Sign is important! Be careful.
You just confused me :-;

Change in temp is positive if we go up, so you get a positive value, hence gaining energy, so the positive sign should be gaining it?
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Eimmanuel
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(Original post by Callicious)
You just confused me :-;
Sorry that the explanation is not easy to understand. This sign issue is confusing at first if no thought is given to it.


(Original post by Callicious)
Change in temp is positive if we go up, so you get a positive value, hence gaining energy, so the positive sign should be gaining it?
Yes.

Let go back to the problem:

An ice cube of mass 0.008kg at 0 degrees Celsius was placed in water at 15 degrees Celsius in an insulated plastic beaker.The mass of water in the beaker was 0.120kg. After the ice cube had melted, the water was stirred, and its temperature was found to have fallen to 9 degrees Celsius.The specific heat capacity of water is 4200J/Kg degrees Celsius.
Q:Calculate the energy transferred from the water.

Assuming that it is an ideal case or "simplified" case, there is no interaction with the surrounding.

In this problem, the ice cube is gaining thermal energy while the water is losing thermal energy.

Using your proposed method:
Using the formula of E = mcΔT, we can find out the energy lost (transferred from the water); E = energy, m = mass, c = specific heat capacity, the last part = change in temperature.

Assume that I have not misinterpreted your method, the energy transferred from water is
mass of water * specific heat capacity * (final temperature - initial temperature)
= 0.12*4200*(9 - 15)
= -3024 J

Energy transferred from water means that the direction of flow of energy is from water to something (in this case is ice cube).
But the answer is negative which means the direction of flow is opposite.

Next, if we use another method (you also proposed this method), but we are not given the specific latent heat of fusion of ice, looking up online, it is 334kJ/kg.

Energy gained by ice and melted ice is
= mass of ice*specific latent heat of fusion of ice
+ mass of ice*specific heat capacity * (final temperature - initial temperature)
= 0.008*334000 + 0.008*4200*(9 - 0)
=2974 J

Energy gained by the ice cube and melted ice means the direction of flow of energy is from somewhere (in this case is water) to the ice cube and melted ice.
In this case, the answer is positive which means the direction of flow is from water to ice cube and melted ice.

You might think there is some calculation error because
energy transferred from water is NOT EQUAL to energy gained by ice and melted ice.

The issue is the final temperature stated in the problem. You can investigate the issue using the method described in this link.
http://www.solvephysics.com/thermody...problem5.shtml

Change in temperature usually means final temperature - initial temperature.
But it does not always give a positive value. It would give a negative value if the temperature decreases.

Hope it is better.
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Callicious
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(Original post by Eimmanuel)
Sorry that the explanation is not easy to understand. This sign issue is confusing at first if no thought is given to it.




Yes.

Let go back to the problem:

An ice cube of mass 0.008kg at 0 degrees Celsius was placed in water at 15 degrees Celsius in an insulated plastic beaker.The mass of water in the beaker was 0.120kg. After the ice cube had melted, the water was stirred, and its temperature was found to have fallen to 9 degrees Celsius.The specific heat capacity of water is 4200J/Kg degrees Celsius.
Q:Calculate the energy transferred from the water.

Assuming that it is an ideal case or "simplified" case, there is no interaction with the surrounding.

In this problem, the ice cube is gaining thermal energy while the water is losing thermal energy.

Using your proposed method:
Using the formula of E = mcΔT, we can find out the energy lost (transferred from the water); E = energy, m = mass, c = specific heat capacity, the last part = change in temperature.

Assume that I have not misinterpreted your method, the energy transferred from water is
mass of water * specific heat capacity * (final temperature - initial temperature)
= 0.12*4200*(9 - 15)
= -3024 J

Energy transferred from water means that the direction of flow of energy is from water to something (in this case is ice cube).
But the answer is negative which means the direction of flow is opposite.

Next, if we use another method (you also proposed this method), but we are not given the specific latent heat of fusion of ice, looking up online, it is 334kJ/kg.

Energy gained by ice and melted ice is
= mass of ice*specific latent heat of fusion of ice
+ mass of ice*specific heat capacity * (final temperature - initial temperature)
= 0.008*334000 + 0.008*4200*(9 - 0)
=2974 J

Energy gained by the ice cube and melted ice means the direction of flow of energy is from somewhere (in this case is water) to the ice cube and melted ice.
In this case, the answer is positive which means the direction of flow is from water to ice cube and melted ice.

You might think there is some calculation error because
energy transferred from water is NOT EQUAL to energy gained by ice and melted ice.

The issue is the final temperature stated in the problem. You can investigate the issue using the method described in this link.
http://www.solvephysics.com/thermody...problem5.shtml

Change in temperature usually means final temperature - initial temperature.
But it does not always give a positive value. It would give a negative value if the temperature decreases.

Hope it is better.
Good explanation :P
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Eimmanuel
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(Original post by Callicious)
Good explanation :P
Since you are doing A-level physics, you should encounter potential difference. If some thought is given to it, one would realize it is related to some change in potential with a sign difference when it is written as  \Delta V in some texts instead of just V.

In mathematics, when we write this  \Delta V, it means that
 \Delta V = V(\text{final}) - V(\text{initial})

But in physics, potential difference is always calculated as
 p.d. = V(\text{high}) - V(\text{low})
because conventional current always flows from high potential to low potential, in this way p.d. is positive.

For  \Delta V, it means that
 \Delta V = V(\text{low}) - V(\text{high})
which has a negative sign but the people who use  \Delta V is expressing it as
 \Delta V = V(\text{high}) - V(\text{low})
which is not really correct.

There are some of these confusing sign issues in physics that are making physics confusing especially when classroom teachers are NOT explaining them.

Again, if you find it confusing, let me know.
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Callicious
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(Original post by Eimmanuel)
Since you are doing A-level physics, you should encounter potential difference. If some thought is given to it, one would realize it is related to some change in potential with a sign difference when it is written as  \Delta V in some texts instead of just V.

In mathematics, when we write this  \Delta V, it means that
 \Delta V = V(\text{final}) - V(\text{initial})

But in physics, potential difference is always calculated as
 p.d. = V(\text{high}) - V(\text{low})
because conventional current always flows from high potential to low potential, in this way p.d. is positive.

For  \Delta V, it means that
 \Delta V = V(\text{low}) - V(\text{high})
which has a negative sign but the people who use  \Delta V is expressing it as
 \Delta V = V(\text{high}) - V(\text{low})
which is not really correct.

There are some of these confusing sign issues in physics that are making physics confusing especially when classroom teachers are NOT explaining them.

Again, if you find it confusing, let me know.
I'm fine with all the potential signage, but it just seems as though you confused me even more with regard to the thermal physics signage. My book cleared it up a notch though
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