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
heat gained by the Al + heat lost by the water = 0\
heat lost by water = q
q = mass water x specific heat water x (Tfinal-Tinitial).
To find the quantity of heat energy absorbed by the piece of aluminum, we can use the formula:
Q = mcΔT
Where:
Q = heat energy
m = mass of the aluminum
c = specific heat capacity of aluminum
ΔT = change in temperature
First, let's convert the mass of the aluminum to kilograms:
m = 22.0 g = 0.022 kg
Next, we need to find the specific heat capacity of aluminum. The specific heat capacity of a substance is the amount of heat energy required to raise the temperature of 1 gram of the substance by 1 degree Celsius.
The specific heat capacity of aluminum is approximately 0.90 J/g°C.
Now let's calculate the change in temperature:
ΔT = final temperature - initial temperature
ΔT = 77.0°C - 0.0°C
ΔT = 77.0°C
Finally, we can substitute all the values into the formula to find the quantity of heat energy absorbed:
Q = (0.022 kg)(0.90 J/g°C)(77.0°C)
Q ≈ 15.51 J
Therefore, the piece of aluminum absorbed approximately 15.51 Joules of heat energy when dropped into the beaker of water.