How much heat must be added to 50.0g of aluminum metal at 25 degrees C to raise its temperature to 125 degrees C?
I have about 20 of these left so if you can do this and work it out and show me how you did it it would be great so I can use it as a guide for the next 19. Thanks.
q = mass x specific heat x (Tf - Ti).
q = 50.0 g x specific heat Al x (125-25).
You will need to look up the specific heat of Al metal. The units should be in Joules/gram. If the units are in J/mol, then you must change the 50.0 g Al to mols Al by dividing 50 by atomic mass Al. Tf is the final T and Ti is initial T in these calculations. You also may have some change of state (e.g. solid to liquid (melting) or liquid to gas (boiling) problems. These are done by
q = mass x heat fusion (for melting)
q = mass x heat of vaporizaton (for boiling). Of course they may be the reverse, also, (opposite of boiling is condensation and opposite of melting is freezing) but those are done exactly the same except the sign is reversed. Heat goes in (plus sign) for melting or boiling and heat goes out (negative sign) for freezing or condensation. Good luck.
ty or this. I tried this al land my answer came to q=24,000 joules. Is this correct...
I looked up Al and found the specific heat to be 0.900 J/g*C.
So 50.0 x 0.900 x 100 = 4500 J.
To calculate the amount of heat required to raise the temperature of a substance, we can use the specific heat capacity equation:
q = mcΔT
Where:
q = heat added (in Joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in J/g°C)
ΔT = change in temperature (in °C)
First, let's determine the specific heat capacity of aluminum. The specific heat capacity of aluminum is 0.897 J/g°C.
Next, we'll substitute the given values into the equation:
m = 50.0g
ΔT = 125°C - 25°C = 100°C
c = 0.897 J/g°C
q = (50.0g)(0.897 J/g°C)(100°C)
Now, let's calculate:
q = (50.0)(0.897)(100)
q = 4485 J
Therefore, the amount of heat that must be added to 50.0g of aluminum metal at 25°C to raise its temperature to 125°C is 4485 Joules.
You can use this approach as a guide for solving similar problems.