If 5.40 kcal of heat is added to 1.00 kg of water at 100C, how much steam at 100C is produced? sHOW ALL CALCULATIONS LEADING TO AN ANSWER. Q=m*Hf 5.4=m*80 m=5.40/80=0.0675kg
You have the wrong formula and wrong data. Hf is heat fusion; you want Hvap. Hf is 80 cal/g; Hvap is 540 cal/g
Q = m*Hvap
Hvap = 540 cal/g
To solve this problem, we need to use the equation Q = m * Hf.
Q represents the amount of heat added, m represents the mass of the water, and Hf represents the heat of fusion.
Given:
Q = 5.40 kcal
m = 1.00 kg
First, we need to determine the heat of fusion (Hf) for water. The heat of fusion is the amount of energy required to convert a substance from a solid to a liquid at its melting point. For water, the heat of fusion is 80 kcal/kg.
Now, we can substitute the values into the equation and solve for the mass of steam (m):
Q = m * Hf
5.40 kcal = m * 80 kcal/kg
Rearranging the equation:
m = 5.40 kcal / 80 kcal/kg
m = 0.0675 kg
Therefore, 0.0675 kg of steam at 100°C is produced.