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physics

A 1.50kg lump of aluminium [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?
Original post by marz1234m
A 1.50kg lump of aluminium [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?
Reply 2
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
Reply 3
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
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?
Reply 5
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
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)
Reply 7
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
Original post by marz1234m
A 1.50kg lump of aluminium [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.

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