If 200 cal heat energy is required 40 grams of liquid to its gaseous state then the latent heat of vapourization of it is
200 cal/40 g = ? cal/g
To calculate the latent heat of vaporization (L), you need to use the formula:
Q = mL
where Q is the heat energy, m is the mass, and L is the latent heat of vaporization.
Given information:
Q = 200 cal
m = 40 g
Step 1: Convert the mass from grams to kilograms:
m = 40 g = 0.04 kg
Step 2: Convert the heat energy from calories to joules:
1 calorie = 4.184 Joules
Q = 200 cal * 4.184 J/cal = 836.8 J
Step 3: Rearrange the formula to solve for L:
L = Q / m
Step 4: Calculate the latent heat of vaporization:
L = 836.8 J / 0.04 kg ≈ 20920 J/kg
Therefore, the latent heat of vaporization of the liquid is approximately 20920 J/kg.
To calculate the latent heat of vaporization, we need to use the formula:
latent heat of vaporization = heat energy / mass
Given that the heat energy required is 200 cal and the mass is 40 grams, we can substitute these values into the formula:
latent heat of vaporization = 200 cal / 40 g
To simplify the calculation, we need to convert calories to joules. 1 calorie is equal to approximately 4.184 joules, so:
latent heat of vaporization = (200 cal * 4.184 J/cal) / 40 g
Now we can calculate the value:
latent heat of vaporization = 8.368 J/g
Therefore, the latent heat of vaporization of the liquid is approximately 8.368 J/g.