How many grams of liquid CF2Cl2 are needed to cool 156 g of water from 42.9 to 20.5 °C? The specific heat of water is 4.184 J/(g

Question

2 years ago

15

·°C).

1 answer:

SashulF [63] · Answer 1 · 2023-04-01T08:29:33+03:00

To cool 156 g of water from 42.9 °C to 20.5 °C, 101 g of CF₂Cl₂ are required.

CF₂Cl₂ is a refrigerant. When it is evaporated, it absorbs heat from water, which cools.

<h3>What is evaporation?</h3>

Evaporation is a type of vaporization that occurs on the surface of a liquid as it changes into the gas phase.

We will use the following expression.

Qw = c × m × ΔT = (4.184 J/g.°C) × 156 g × (20.5 °C - 42.9 °C)

Qw = -14.6 kJ

where,

If water releases 14.6 kJ of heat, CF₂Cl₂ absorbs 14.6 kJ of heat (Qr = 14.6 kJ).

We will use the following expression.

Qr = ΔH°evap × m

m = Qr/ΔH°evap = 14.6 kJ / (0.144 kJ/g) = 101 g

where,

To cool 156 g of water from 42.9 °C to 20.5 °C, 101 g of CF₂Cl₂ are required.

Learn more about evaporation here: brainly.com/question/25310095