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Desperately need someone to help me answer this question!

The initial temperature of a mixture was +20 degrees. The mixture froze at -1.5 degrees.

A total of 165 KJ of internal energy was transferred from the mixture to cool and freeze it.

specific latent heat= 3500 J/KJ degrees

specific latent heat of fusion of the mixture=255 000 J/Kg

Calculate the mass of the mixture.

Give your answer to 2 s.f

The initial temperature of a mixture was +20 degrees. The mixture froze at -1.5 degrees.

A total of 165 KJ of internal energy was transferred from the mixture to cool and freeze it.

specific latent heat= 3500 J/KJ degrees

specific latent heat of fusion of the mixture=255 000 J/Kg

Calculate the mass of the mixture.

Give your answer to 2 s.f

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#2

What have you tried, what do you know?

For cooling, think about specific latent heat. For mass,specific-latent-heat,temp-change,energy-transfer, you have

For freezing, think about latent heat of fusion , i.e.

For cooling, think about specific latent heat. For mass,specific-latent-heat,temp-change,energy-transfer, you have

For freezing, think about latent heat of fusion , i.e.

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(Original post by

What have you tried, what do you know?

For cooling, think about specific latent heat. For mass,specific-latent-heat,temp-change,energy-transfer, you have

For freezing, think about latent heat of fusion , i.e.

**Callicious**)What have you tried, what do you know?

For cooling, think about specific latent heat. For mass,specific-latent-heat,temp-change,energy-transfer, you have

For freezing, think about latent heat of fusion , i.e.

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#4

(Original post by

I was just confused on when the energy transferred is from the freezing process and cooling. As how do you put that into one of the equations when its from both of them...

**Scarlett.sully**)I was just confused on when the energy transferred is from the freezing process and cooling. As how do you put that into one of the equations when its from both of them...

Would it help if I told you that you had

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(Original post by

They give you 165 kJ and you have the from each of the processes individually. Both occur, and you can just assume the mass of the substance doesn't change.

Would it help if I told you that you had

**Callicious**)They give you 165 kJ and you have the from each of the processes individually. Both occur, and you can just assume the mass of the substance doesn't change.

Would it help if I told you that you had

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#6

(Original post by

Yes that defiantly helped me understand the concept, however how do you guess or workout the mass if the energy transferred if the number is from both equations. so if you put it into the SHC, for example, it will also include the energy transferred is SLH, therefore not giving you the correct mass

**Scarlett.sully**)Yes that defiantly helped me understand the concept, however how do you guess or workout the mass if the energy transferred if the number is from both equations. so if you put it into the SHC, for example, it will also include the energy transferred is SLH, therefore not giving you the correct mass

and

for cooling/fusing respectively. Their sum (as mentioned) is the total heat transferred from the substance

I might have been a bit inexact with how I labelled my EQ's for you ( is

*different*for fusing/cooling, equation-wise... I was just using the same to be lazy and assumed in context it'd be fine)

and you can redistribute via

where you can rearrange for .

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(Original post by

Mmm... take both the equations I gave, i.e.

and

for cooling/fusing respectively. Their sum (as mentioned) is the total heat transferred from the substance

I might have been a bit inexact with how I labelled my EQ's for you ( is

and you can redistribute via

where you can rearrange for .

**Callicious**)Mmm... take both the equations I gave, i.e.

and

for cooling/fusing respectively. Their sum (as mentioned) is the total heat transferred from the substance

I might have been a bit inexact with how I labelled my EQ's for you ( is

*different*for fusing/cooling, equation-wise... I was just using the same to be lazy and assumed in context it'd be fine)and you can redistribute via

where you can rearrange for .

165000J

------------- to find the mass at the specific heat capacity

3500 x 21.5

165000

---------------- to find the mass at the Specific latent heat

255000

and then subtract them from each other?

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#8

(Original post by

OHHhhhh ok so you would do

165000J

------------- to find the mass at the specific heat capacity

3500 x 21.5

165000

---------------- to find the mass at the Specific latent heat

255000

and then subtract them from each other?

**Scarlett.sully**)OHHhhhh ok so you would do

165000J

------------- to find the mass at the specific heat capacity

3500 x 21.5

165000

---------------- to find the mass at the Specific latent heat

255000

and then subtract them from each other?

**Callicious**)

Mmm... take both the equations I gave, i.e.

and

for cooling/fusing respectively. Their sum (as mentioned) is the total heat transferred from the substance

I might have been a bit inexact with how I labelled my EQ's for you ( is

*different*for fusing/cooling, equation-wise... I was just using the same to be lazy and assumed in context it'd be fine)

and you can redistribute via

where you can rearrange for .

Here...

You need to

*rearrange*the last equation.

Last edited by Callicious; 1 month ago

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(Original post by

There isn't any mass loss...

The first is an expression of the heat transferred from the mixture when cooling/heating it. The second is an expression for the heat transferred from the mixture when it freezes/melts. Their sum is the third expression, which is the quantity you've been given, 165,000J.

Here...

You need to

**Callicious**)There isn't any mass loss...

The first is an expression of the heat transferred from the mixture when cooling/heating it. The second is an expression for the heat transferred from the mixture when it freezes/melts. Their sum is the third expression, which is the quantity you've been given, 165,000J.

Here...

You need to

*rearrange*the last equation.and then adding on 255000 = 330250

would you then divide by the energy transferred?

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#10

(Original post by

mmm ok so if i rearanged this i would get 3500 x 21.5 =75250

and then adding on 255000 = 330250

would you then divide by the energy transferred?

**Scarlett.sully**)mmm ok so if i rearanged this i would get 3500 x 21.5 =75250

and then adding on 255000 = 330250

would you then divide by the energy transferred?

Ideally you'd address the final step with one big sum dumped into your calculator, which avoids dealing with numbers along the way

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(Original post by

Other way round, divide energy transferred by that quantity. If you rearrange that equation I gave you, you'd get

Ideally you'd address the final step with one big sum dumped into your calculator, which avoids dealing with numbers along the way

**Callicious**)Other way round, divide energy transferred by that quantity. If you rearrange that equation I gave you, you'd get

Ideally you'd address the final step with one big sum dumped into your calculator, which avoids dealing with numbers along the way

------------- = 0.5 kg

330250

I put it into the calculation and it works! i wanna thank you so much for you patience, this has helped me so much haha

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#12

Is the change in energy not -21.5 rather than 21.5, If I'm wrong please explain as I'm a little confused

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#13

(Original post by

Is the change in energy not -21.5 rather than 21.5, If I'm wrong please explain as I'm a little confused

**kemblespotter**)Is the change in energy not -21.5 rather than 21.5, If I'm wrong please explain as I'm a little confused

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#14

(Original post by

yes the change in energy is -21.5 because you have to do (-1.5) - 20 to get the change in temp.

**Uwais Latona**)yes the change in energy is -21.5 because you have to do (-1.5) - 20 to get the change in temp.

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