Physics GCSE

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

Reply 1

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

mc_s\Delta{\Theta} = \Delta{E}

For freezing, think about latent heat of fusion

q_f

, i.e.

q_f{m} = \Delta{E}

Reply 2

Scarlett.sully

Original post by 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

mc_s\Delta{\Theta} = \Delta{E}

For freezing, think about latent heat of fusion

q_f

, i.e.

q_f{m} = \Delta{E}

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...

Reply 3

Callicious

Original post by 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...

They give you 165 kJ and you have the

\Delta{E}

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

165\textrm{kJ} = \Delta_\Sigma{E} = \Delta_{Freezing}E + \Delta_{Cooling}E

Reply 4

Scarlett.sully

Original post by Callicious

They give you 165 kJ and you have the

\Delta{E}

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

165\textrm{kJ} = \Delta_\Sigma{E} = \Delta_{Freezing}E + \Delta_{Cooling}E

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

Reply 5

Callicious

Original post by Scarlett.sully

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

\Delta{E} = mc_s\Delta\Theta

and

\Delta{E} = q_f{m}

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

\Delta{E}_\Sigma = mc_s\Delta\Theta + q_f{m}

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

\Delta{E}

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

m

via

\Delta{E}_\Sigma = m\left(c_s\Delta\Theta + q_f\right)

where you can rearrange for

m

Reply 6

Scarlett.sully

Original post by Callicious

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

\Delta{E} = mc_s\Delta\Theta

and

\Delta{E} = q_f{m}

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

\Delta{E}_\Sigma = mc_s\Delta\Theta + q_f{m}

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

\Delta{E}

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

m

via

\Delta{E}_\Sigma = m\left(c_s\Delta\Theta + q_f\right)

where you can rearrange for

m

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?

Reply 7

Callicious

Original post by Scarlett.sully

There isn't any mass loss...

Original post by Callicious

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

\Delta{E} = mc_s\Delta\Theta

and

\Delta{E} = q_f{m}

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

\Delta{E}_\Sigma = mc_s\Delta\Theta + q_f{m}

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

\Delta{E}

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

m

via

\Delta{E}_\Sigma = m\left(c_s\Delta\Theta + q_f\right)

where you can rearrange for

m

The first

\Delta{E}

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...

(\Delta{E})_{COOLING} = mc_s\Delta\Theta

(\Delta{E})_{FREEZING} = q_f{m}

(\Delta{E})_{TOTAL=\Sigma} = mc_s\Delta\Theta + q_f{m} = m(c_s\Delta\Theta + q_f) = 165\textrm{kJ}

You need to rearrange the last equation.

(edited 2 years ago)

Reply 8

Scarlett.sully

Original post by Callicious

There isn't any mass loss...

The first

\Delta{E}

(\Delta{E})_{COOLING} = mc_s\Delta\Theta

(\Delta{E})_{FREEZING} = q_f{m}

(\Delta{E})_{TOTAL=\Sigma} = mc_s\Delta\Theta + q_f{m} = m(c_s\Delta\Theta + q_f) = 165\textrm{kJ}

You need to rearrange the last equation.

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?

Reply 9

Callicious

Original post by 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?

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

\displaystyle m = \frac{\left(\Delta{E}\right)_{TOTAL=\Sigma}}{c_s\Delta\Theta + q_f}

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

Reply 10

Scarlett.sully

Original post by Callicious

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

\displaystyle m = \frac{\left(\Delta{E}\right)_{TOTAL=\Sigma}}{c_s\Delta\Theta + q_f}

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

i get iiittt so the final calculation is 165000
------------- = 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 :wink: