How much energy is required to heat 32.5 g of water from 34°C to 75°C
1 answer:
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Answer:
14,448 J of heat would it take to completely vaporize 172 g of this liquid at its boiling point.
Explanation:
The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to Q = m*L, where L is called the latent heat of the substance and depends on the type of phase change.
During the evaporation process, a substance goes from a liquid to a gaseous state and needs to absorb a certain amount of heat from its immediate surroundings, which results in its cooling. The heat absorbed is called the heat of vaporization.
So, it is called "heat of vaporization", the energy required to change 1 gram of substance from a liquid state to a gaseous state at the boiling point.
In this case, being:
- Q=?
- m= 172 g
- L= 84
and replacing in the expression Q = m*L you get:
Q=172 g*84
Q=14,448 J
<u><em>14,448 J of heat would it take to completely vaporize 172 g of this liquid at its boiling point.</em></u>