A 22.0 g piece of aluminum at 0.0°C is dropped into a beaker of water. The temperature of the water drops from 92.0°C to 77.0°C. What quantity of heat energy did the piece of aluminum absorb?
Please Help, I'm so lost :(
Thank you
See your post above. I missed this one when I checked. Sorry you had to post twice.
To calculate the quantity of heat energy absorbed by the piece of aluminum, you can use the formula:
Q = m * c * ΔT
Where:
Q is the heat energy absorbed or released,
m is the mass of the substance (in this case, the aluminum),
c is the specific heat capacity of the substance (in this case, the specific heat capacity of aluminum), and
ΔT is the change in temperature.
First, let's find the value of m, the mass of the aluminum. Given that the mass is 22.0 g, we can use this value directly.
Second, we need to find the specific heat capacity of aluminum. The specific heat capacity of aluminum is approximately 0.897 J/g°C.
Now, we can calculate the quantity of heat energy absorbed by the aluminum:
Q = m * c * ΔT
Where:
m = 22.0 g,
c = 0.897 J/g°C,
ΔT = (final temperature - initial temperature)
ΔT = (77.0°C - 0.0°C) = 77.0°C
Q = 22.0 g * 0.897 J/g°C * 77.0°C
Calculating this, we get:
Q = 1511.6 J
Therefore, the piece of aluminum absorbed approximately 1511.6 J of heat energy.