question 3 of 10 how much energy is required to vaporize 1.5 kg of aluminum? (refer to table of latent heat…

Explanation:

Step1: Recall the formula for heat of vaporization

$Q = mL_v$, where $Q$ is the heat energy, $m$ is the mass, and $L_v$ is the latent - heat of vaporization. The latent heat of vaporization of aluminum, $L_v=4.73\times 10^{6}\ J/kg$.

Step2: Substitute the given values

$m = 1.5\ kg$ and $L_v=4.73\times 10^{6}\ J/kg$ into the formula $Q = mL_v$. So, $Q=1.5\ kg\times4.73\times 10^{6}\ J/kg = 7.095\times 10^{6}\ J$.

Step3: Convert to kilojoules

Since $1\ kJ = 1000\ J$, then $Q=\frac{7.095\times 10^{6}\ J}{1000}=7095\ kJ$.

Answer:

B. 7095 kJ