q=MCAT

[jsmith1]

Hey guys I had a few questions about this equation. Does the q stand for the change in energy that must be provided (or removed) to change the temp of a substance? How would you conceptually tackle a problem that involves combining a hot substance with a cool that come together to an equilibrium temperature?

50g piece of an unknown metal at 50C is added to 50g H2O at 0C and the mixture reaches thermal equil at 10C. What is the specific heat of the metal?

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[FeinMS]

conceptually, heat is conserved in a closed system. So heat lost by one object equals heat gained by another. q_Al=-q_H2O Note the negative sign.

 

[Gauss44]

Hey guys I had a few questions about this equation. Does the q stand for the change in energy that must be provided (or removed) to change the temp of a substance? How would you conceptually tackle a problem that involves combining a hot substance with a cool that come together to an equilibrium temperature?

50g piece of an unknown metal at 50C is added to 50g H2O at 0C and the mixture reaches thermal equil at 10C. What is the specific heat of the metal?

I don't know if this is correct, but I would go:

water
q=50*4*10
q=2000

unknown metal
2000=50*C*40
C=1

(q=energy, C=specific heat or heat capacity)

 

[haloalkane]

First, you take the given numbers and solve Q for water. You should be able to do it because "c" for H2O is known (you can look it up, or it should be given). Now that you have found "Q" re-arrange the equation to c=q/mt to solve for "unknown". Plug them in and see what you get!

 

[Synapsis]

Remember that heat (q) is just a form of energy transfer. There's two forms that we really care about for the MCAT, and those are heat and work. Heat is energy transfer based on differential temperatures. Work is essentially any other energy transfer.

So since heat is transfer of energy between two substances, then if one loses energy, that same energy is gained by the second. So if we put a hot metal in water, the hot metal will lose some energy in the form of heat transfer right? That same energy, q, is acquired by the water, which is at a lower temperature. So, you can set the q=mc(delta)T equation for the metal equal to the q = mc(delta)T equation for the liquid, since the q is the same in both cases.

MmetalCmetal(delta)T = MwaterCwater(delta)T

Now the system reaches thermal equilibrium at 10 degrees Celsius, so the metal changes from 50 C to 10C, meaning (delta)T = 50 - 10 = 40 C. The water changes from 0 C to 10 C so (delta)T for the water is 10 C. Using the heat capacity of water, the mass of water, the mass of metal, and the change in temperatures, it just becomes a plug and chug.

 

[thun]

Heat gained = - Heat lost
Q cold = -Q hot


C unknown = [ (.05) x (4190) x (10-0)]/[-(.05) x (10 - 50)]