Respuesta :

dsdrajlin

Answer:

1.62 × 10⁴ cal

Explanation:

Step 1: Given and required data

  • Mass of water (m): 30.0 g
  • Latent heat of vaporization of water (ΔH°vap): 540 cal/g

Step 2: Calculate the heat change (Q) for vaporization of 30.0g of water at 100 °C

The vaporization is a phase change in which water passes from the liquid state to the gaseous state. We can calculate the heat required using the following expression.

Q = ΔH°vap × m

Q = 540 cal/g × 30.0 g

Q = 1.62 × 10⁴ cal

mickymike92

Based on calculation of Heat change, the heat change involved is 1.62 × 10⁴ cal.

What is the heat of vaporization of a liquid?

Heat of vaporization of a liquid is the quantity of Heat required to convert a unit mass of a liquid to vapor at its boiling point.

The heat required is given by the given formula:

  • Q = ΔH°vap × m

where:

  • Q is quantity of Heat involved
  • m is mass of liquid
  • ΔH°vap is the heat of vaporization of the liquid

How to calculate the heat change

From the data provided:

Mass of water (m): 30.0 g

Latent heat of vaporization of water (ΔH°vap): 540 cal/g

Using the formula of Heat change

Q = ΔH°vap × m

Q = 540 cal/g × 30.0 g

Q = 1.62 × 10⁴ cal

Therefore, the heat change involved is 1.62 × 10⁴ cal.

Learn more about heat of vaporization at: https://brainly.com/question/11475740

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