# physics

Watch
Announcements
#1
A 1.50kg lump of aluminium [c=910J/(kg ∘C)]at 100, degrees is dropped into a beaker containing 1.00kg of water at 20.0 ∘C.
The specific heat capacity of pure water is 4200J/(kg ∘C).

Assuming the system is well insulated, what temperature will the aluminium and water be at thermal equilibrium?
0
8 months ago
#2
(Original post by marz1234m)
A 1.50kg lump of aluminium [c=910J/(kg ∘C)]at 100, degrees is dropped into a beaker containing 1.00kg of water at 20.0 ∘C.
The specific heat capacity of pure water is 4200J/(kg ∘C).

Assuming the system is well insulated, what temperature will the aluminium and water be at thermal equilibrium?
Why are you not posting this in the physics forum?

The final temperature of both the aluminium and the water will be the same (that's the question).
You know the energy change in the aluminium must equal the energy change in the water.
Can you see how to proceed?
0
#3
(Original post by charco)
Why are you not posting this in the physics forum?

The final temperature of both the aluminium and the water will be the same (that's the question).
You know the energy change in the aluminium must equal the energy change in the water.
Can you see how to proceed?
i think so , i'll have another go ,thank you
0
#4
do i add the change in energies and half and then divide by the total mass if so which specific heat capacity would i divide by
i
0
8 months ago
#5
(Original post by marz1234m)
do i add the change in energies and half and then divide by the total mass if so which specific heat capacity would i divide by
i
No.

Like I said the energy lost by the aluminium = the energy gained by the water.

Energy lost by the aluminium = m(Al) x c(Al) x ΔT

and ΔT = 100 - T(final)

Energy gained by the water = m(w) x c(Al) x ΔT

and in this case ΔT = T(final) - 20

can you go on now?
0
#6
(Original post by charco)
No.

Like I said the energy lost by the aluminium = the energy gained by the water.

Energy lost by the aluminium = m(Al) x c(Al) x ΔT

and ΔT = 100 - T(final)

Energy gained by the water = m(w) x c(Al) x ΔT

and in this case ΔT = T(final) - 20

can you go on now?
i understand it a lot more but i still dont know how to proceed
0
8 months ago
#7
(Original post by marz1234m)
i understand it a lot more but i still don't know how to proceed
Energy lost by the aluminium = m(Al) x c(Al) x ΔT

and ΔT = 100 - T(final)

Energy gained by the water = m(w) x c(Al) x ΔT

and in this case ΔT = T(final) - 20

------------------------------------------------------------------------------- the energy is equal therefore

m(Al) x c(Al) x (100 - T(final)) = m(w) x c(Al) x (T(final) - 20)

Now substitute values and solve for T(final)
0
#8
(Original post by charco)
Energy lost by the aluminium = m(Al) x c(Al) x ΔT

and ΔT = 100 - T(final)

Energy gained by the water = m(w) x c(Al) x ΔT

and in this case ΔT = T(final) - 20

------------------------------------------------------------------------------- the energy is equal therefore

m(Al) x c(Al) x (100 - T(final)) = m(w) x c(Al) x (T(final) - 20)

Now substitute values and solve for T(final)
thank you for your help really appreciate it
0
8 months ago
#9
(Original post by marz1234m)
A 1.50kg lump of aluminium [c=910J/(kg ∘C)]at 100, degrees is dropped into a beaker containing 1.00kg of water at 20.0 ∘C.
The specific heat capacity of pure water is 4200J/(kg ∘C).

Assuming the system is well insulated, what temperature will the aluminium and water be at thermal equilibrium?
you do a subject as mind-boggling as physics, big respect to you.
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Feeling behind at school/college? What is the best thing your teachers could to help you catch up?

Extra compulsory independent learning activities (eg, homework tasks) (13)
7.34%
Run extra compulsory lessons or workshops (29)
16.38%
Focus on making the normal lesson time with them as high quality as possible (29)
16.38%
Focus on making the normal learning resources as high quality/accessible as possible (25)
14.12%
Provide extra optional activities, lessons and/or workshops (48)
27.12%
Assess students, decide who needs extra support and focus on these students (33)
18.64%

View All
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.