Respuesta :

dsdrajlin

Answer:

-2.79 × 10³ cal

Explanation:

Step 1: Given data

  • Mass of water (m): 35.0 g
  • Latent heat of fusion of water (L): -79.7 cal/g

Step 2: Calculate the heat required to freeze 35.0 g of water

We have 35.0 g of liquid water and we want to freeze it, that is, to convert it in 35.0 g of ice (solid water), at 0 °C (melting point). We can calculate the heat (Q) that must be released using the following expression.

Q = L × m

Q = -79.7 cal/g × 35.0 g

Q = -2.79 × 10³ cal

samueladesida43

The heat in calories required to freeze 35.0g of water of water is 2.79 × 10³ cal.

HOW TO CALCULATE HEAT:

  • The heat of a water in the mass of water can be calculated as follows:

Q = L × m

Where;

  1. Q = quantity of heat (calories)
  2. L = latent heat of fusion of water
  3. m = mass of water

According to this question;

  1. Mass of water (m): 35.0 g
  2. Latent heat of fusion of water (L): -79.7 cal/g

Q = L × m

Q = -79.7 cal/g × 35.0 g

Q = -2.79 × 10³ calories

Therefore, the heat in calories required to freeze 35.0g of water of water is 2.79 × 10³ cal.

Learn more about how to calculate heat at: https://brainly.com/question/21122443