A block of aluminum with a volume of 96.6 cm3 absorbs 66.6 J of heat. If its initial temperature was 32.5 °C, what is its final temperature? (density of aluminum = 2.70 g/cm3
find the mass of the block: density*volume
heat= mass*specificheatAl*(Tf-32.5)
look up the specificheat for Al, and solve for the final temp.
To calculate the final temperature of the block of aluminum, we can use the formula for heat transfer:
Q = mcΔT
Where:
Q is the heat transferred (in joules),
m is the mass of the aluminum block (in grams),
c is the specific heat capacity of aluminum (in J/g°C),
ΔT is the change in temperature (in °C).
First, let's calculate the mass of the aluminum block using its volume and density:
Density = mass / volume
mass = density × volume
mass = 2.70 g/cm3 × 96.6 cm3
mass ≈ 260.82 grams
Next, we need to find the specific heat capacity of aluminum. The specific heat capacity of aluminum is typically around 0.897 J/g°C.
Now we can rearrange the formula to solve for the change in temperature:
ΔT = Q / (m × c)
ΔT = 66.6 J / (260.82 g × 0.897 J/g°C)
ΔT ≈ 0.259 °C
Lastly, to find the final temperature, we add the change in temperature to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 32.5 °C + 0.259 °C
Final temperature ≈ 32.76 °C
Therefore, the final temperature of the aluminum block is approximately 32.76 °C.