Question

A. 207 kJ B. 4730 kJ C. 9460 kJ D. 414 kJ

Answer 1

Answer:

$\boxed {\boxed {\sf C. \ 9460 \ kJ}}$

Explanation:

The formula we must use is given to us:

$q=mL_{vapor}$

q is the energy, m is the mass, and L(vapor) is the latent heat of vaporization.

The energy is what we calculate and the mass is 2 kilograms. We need to find the latent heat of vaporization, which is on the table.

• We know the sample is copper.
• Find that element on the table, then the third box tells us it's latent heat of vaporization is 4730 kJ/kg

Now we know:

$m= 2 \ kg \\L_{vapor}=4730 \ kJ/kg$

Substitute the values into the formula.

$q=(2 \ kg ) * (4730 \ kJ/kg)$

Multiply. Note that the kilograms (kg) will cancel each other out.

$q= 2* 4730 \ kJ$

$q= 9460 \ kJ$

9460 kilojoules are required to vaporize 2 kilograms of copper.