# (PHYSICS) Dear lord someone help meWatch

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#1
I hate isaac physics so damn much. Please help i've been stuck on the question quite literally all day
https://isaacphysics.org/questions/c...lls_book_ch_g4

Just part G4.10 !!

I think i'm close but i'm losing my sanity
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1 year ago
#2
How much energy is required to heat the ice, melt it and then heat it to a temperature T?

Home much energy is lost for the water to cool to a temperature T?

Equate the two and solve. If there is no solution, the ice doesn't melt, so redo assuming that the final state of the combination is solid.
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#3
(Original post by RogerOxon)
How much energy is required to heat the ice, melt it and then heat it to a temperature T?

Home much energy is lost for the water to cool to a temperature T?

Equate the two and solve. If there is no solution, the ice doesn't melt, so redo assuming that the final state of the combination is solid.
Oh man I'm losing it. Can you see what's wrong with that second line I wrote? Just ignore the rest I have like three pages of mistakes at this point
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#4
(Original post by I'msoPi)
Oh man I'm losing it. Can you see what's wrong with that second line I wrote? Just ignore the rest I have like three pages of mistakes at this point
Maybe heating the ice is supposed to be 15 +Tf?
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1 year ago
#5
Sorry, Isaac at physics.
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#6
(Original post by Luminescence0)
Sorry, Isaac at physics.
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#7
(Original post by I'msoPi)
I hate isaac physics so damn much. Please help i've been stuck on the question quite literally all day
https://isaacphysics.org/questions/c...lls_book_ch_g4

Just part G4.10 !!

I think i'm close but i'm losing my sanity
heart breaks a little everytime it says incorrect
0
1 year ago
#8
Ice gains energy as it heats up from -15 to 0 degrees.
It's safe to assume that all of the ice then melts (due to the 59 degree starting temp. of "beaker water", and its greater mass). This also requires energy.
The melted "ice water" will then heat up to a certain temperature, T, where the "ice water" and "beaker water" are in thermal equilibrium.
All 3 of these processes require thermal energy from the "beaker water", as it cools from 59 to T degrees.
You can form a single equation involving T using this information.
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#9
(Original post by dan1yar18)
Ice gains energy as it heats up from -15 to 0 degrees.
It's safe to assume that all of the ice then melts (due to the 59 degree starting temp. of "beaker water", and its greater mass). This also requires energy.
The melted "ice water" will then heat up to a certain temperature, T, where the "ice water" and "beaker water" are in thermal equilibrium.
All 3 of these processes require thermal energy from the "beaker water", as it cools from 59 to T degrees.
You can form a single equation involving T using this information.
Would you mind breaking it up for me further. So is it

Energy melting the ice + energy to heat ice to water + additional energy to heat ice water = energy to heat the combination of ice water and beaker water

I've been on this for hours so my brain is fried sorry
0
1 year ago
#10
(Original post by I'msoPi)
Would you mind breaking it up for me further. So is it

Energy melting the ice + energy to heat ice to water + additional energy to heat ice water = energy to heat the combination of ice water and beaker water

I've been on this for hours so my brain is fried sorry

The ice needs energy to:
- Heating from -15C to 0C
- Melting
- Heating from 0C to T

The water releases energy:
- Cooling from 59C to T

Set the two equal.
1
1 year ago
#11
(Original post by I'msoPi)
Would you mind breaking it up for me further. So is it

Energy melting the ice + energy to heat ice to water + additional energy to heat ice water = energy to heat the combination of ice water and beaker water

I've been on this for hours so my brain is fried sorry
You've got the left side of the equation right... but think about where all of that energy is gained from...
What is happening is that heat energy is being transferred from the beaker water to the ice/ice water.
This means that the right side of the equation should be the heat energy lost by only the beaker water.

That aside, in your previous working you have combined the masses of ice water and beaker water using mcΔT - this cannot be done as they are at different temperatures so will have different changes in temperature.

You're very close, but I'd say come back to it another time if you haven't got it yet.
1
1 year ago
#12
(Original post by I'msoPi)
Oh man I'm losing it. Can you see what's wrong with that second line I wrote? Just ignore the rest I have like three pages of mistakes at this point
Your mistake is that the melted ice heats from 0C to T, whereas the water cools from 59C to T. You've counted the melted ice as cooling from 59C.
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#13
(Original post by RogerOxon)
Your mistake is that the melted ice heats from 0C to T, whereas the water cools from 59C to T. You've counted the melted ice as cooling from 59C.
I'm so confused. I took a break from this question but nothing

How does this look to you

(0.35)(3.35x10^5) + (0.35)(2030)(15) + (0.35)(2030)T = 0.61(4180)(59-T) For the right hand side I would've added the masses but someone has corrected me in saying that I shouldn't but i've had a note from isaac physics saying "Incorrect Have you thought about what happens to the ice water after the ice has melted? Please try again."

Just want this to be over with - lost an entire day of revision to one stupid question
0
1 year ago
#14
(Original post by I'msoPi)
I'm so confused. I took a break from this question but nothing

How does this look to you

(0.35)(3.35x10^5) + (0.35)(2030)(15) + (0.35)(2030)T = 0.61(4180)(59-T) For the right hand side I would've added the masses but someone has corrected me in saying that I shouldn't but i've had a note from isaac physics saying "Incorrect Have you thought about what happens to the ice water after the ice has melted? Please try again."

Just want this to be over with - lost an entire day of revision to one stupid question
Almost. You've got the wrong specific heat capacity for water on the LHS. Otherwise, that looks correct.
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#15
(Original post by RogerOxon)
Almost. You've got the wrong specific heat capacity for water on the LHS. Otherwise, that looks correct.
Ah yeah I fixed that but now it says that my significant figures are wrong. I'm very much on the verge of a mental breakdown. I ended up getting 0.6822 whatever which is probably still wrong
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#16
(Original post by RogerOxon)
Almost. You've got the wrong specific heat capacity for water on the LHS. Otherwise, that looks correct.
Mistyped it into the calculator. I Finally have it!!! Bless you everyone that helped
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