How many kilocalories ( Kcal) of heat are needed to vaporize 35.0 grams of water to its vapor at 100 Celsius? Heat of vaporization Of H20=540 calories /1g H2O
To find the total amount of heat needed to vaporize water, we can use the formula:
Heat = Mass × Heat of vaporization
Given:
Mass of water (m) = 35.0 grams
Heat of vaporization of water (Hv) = 540 calories/1g water
First, let's convert the mass of water from grams to kilograms:
Mass of water = 35.0 grams = 35.0 grams × (1 kilogram / 1000 grams) = 0.035 kilograms
Now, we can calculate the heat needed to vaporize the water:
Heat = Mass of water × Heat of vaporization
Heat = 0.035 kilograms × 540 calories/1g water
To convert the result to kilocalories, we divide the value by 1000:
Heat = (0.035 kilograms × 540 calories/1g water) / 1000
Heat = 18.9 kilocalories
Therefore, 35.0 grams of water would require 18.9 kilocalories of heat to vaporize at 100 degrees Celsius.
To calculate the amount of heat needed to vaporize a given amount of water, we can use the equation:
Heat (in kcal) = mass (in grams) × heat of vaporization (in cal/g)
Given:
Mass of water (m) = 35.0 grams
Heat of vaporization of water (Hv) = 540 cal/g
First, let's convert the mass from grams to kilograms:
Mass (m) = 35.0 grams = 0.035 kilograms
Next, we need to convert the heat of vaporization from calories to kilocalories:
Heat of vaporization (Hv) = 540 cal/g = 0.54 kcal/g
Now, we can use the formula to calculate the heat required to vaporize the given mass of water:
Heat (in kcal) = mass (in kilograms) × heat of vaporization (in kcal/g)
Heat (in kcal) = 0.035 kg × 0.54 kcal/g
Heat (in kcal) = 0.0189 kcal
Therefore, approximately 0.0189 kilocalories (Kcal) of heat are needed to vaporize 35.0 grams of water at 100 degrees Celsius.
q = 540 cal/g x 35 g = ? calories (note g calcels and leaves cal)
Then convert to kcal.