To determine how many joules of heat [tex]\( Q \)[/tex] are required to completely evaporate 37.5 grams of boiling water at a temperature of [tex]\( 100^\circ \text{C} \)[/tex], we will use the concept of latent heat of vaporization. The latent heat of vaporization is the amount of heat needed to convert a unit mass of a substance from a liquid to a gas without a change in temperature.
The latent heat of vaporization for water at [tex]\( 100^\circ \text{C} \)[/tex] is given as [tex]\( 2260 \, \text{J/g} \)[/tex]. This means that [tex]\( 2260 \, \text{J} \)[/tex] of energy is required to vaporize 1 gram of water at this temperature.
We can use the following formula to calculate the total heat required:
[tex]\[
Q = m \cdot L
\][/tex]
where:
- [tex]\( Q \)[/tex] is the total heat required (in joules),
- [tex]\( m \)[/tex] is the mass of the water (in grams),
- [tex]\( L \)[/tex] is the latent heat of vaporization (in joules per gram).
Given:
- [tex]\( m = 37.5 \, \text{g} \)[/tex],
- [tex]\( L = 2260 \, \text{J/g} \)[/tex].
Plug in the values into the formula:
[tex]\[
Q = 37.5 \, \text{g} \times 2260 \, \text{J/g}
\][/tex]
Performing the multiplication:
[tex]\[
Q = 84850 \, \text{J}
\][/tex]
Therefore, the total amount of heat needed to completely evaporate 37.5 grams of boiling water at [tex]\( 100^\circ \text{C} \)[/tex] is [tex]\( 84750.0 \, \text{J} \)[/tex].
