Liquid Freon (CCl2F2) is used as a refrigerant. It is circulated inside the cooling coils of older refrigerators or freezers. As it evaporates, it absorbs heat. How much heat can be removed by 2.00 kg of Freon as it evaporates inside the coils of a refrigerator? The heat of vaporization of Freon is 38.6 g/cal.
Please help!
I'm also wondering why no temperature given. I don't understand why it wouldn't be needed.
Thanks
Something is wrong. The heat of vaporization is given in units of cal/gram or cal/mol. .
Oh my apologies! It's 38.6 cal/g
OK. So the heat of vaporization (at whatever temperature that is)is 38.6 cal/gram CH2Cl2. That's why no T is given. How many grams do you have? That's 2 kg or 2,000 grams.
So 38.6 ca/g x 2000 g = ? cal. And you thought this was complicated/
To find the heat that can be removed by 2.00 kg of Freon as it evaporates inside the coils of a refrigerator, we can use the formula:
Heat = Mass × Heat of Vaporization
Given:
Mass of Freon (m) = 2.00 kg
Heat of Vaporization (Hv) = 38.6 g/cal
First, we need to convert the heat of vaporization from grams to kilograms:
38.6 g/cal = 38.6 g / 1000 g/kg = 0.0386 kg/cal
Now we can calculate the heat that can be removed:
Heat = Mass × Heat of Vaporization
Heat = 2.00 kg × 0.0386 kg/cal
Calculating this equation will give us the result:
Heat = 0.0772 kg·cal
Therefore, 2.00 kg of Freon has the capacity to remove approximately 0.0772 kg·cal of heat as it evaporates inside the coils of a refrigerator.