Find mass of liquid water required to absorb 3.94 X e4 of heat energy on boiling. Answer in units of g
3.94E4 what?
mass H2O x heat vap = 3.94E4 whatever.
Solve for mass H2O
To find the mass of liquid water required to absorb a certain amount of heat energy during boiling, we need to use the specific heat capacity of water and the heat energy equation.
The specific heat capacity of water is defined as 4.18 J/(g°C), meaning it takes 4.18 Joules of energy to raise the temperature of 1 gram (g) of water by 1 degree Celsius (°C).
The heat energy equation is given by Q = m * c * ΔT, where Q is the heat energy absorbed or released, m is the mass of the substance, c is the specific heat capacity, and ΔT is the temperature change.
In this case, we want to find the mass (m) of water. We know the heat energy (Q) is 3.94 × 10^4 Joules and the temperature change (ΔT) during boiling is 100°C. We can rearrange the equation to solve for mass (m):
m = Q / (c * ΔT)
Plugging in the given values, we have:
m = 3.94 × 10^4 J / (4.18 J/(g°C) * 100°C)
Simplifying the equation, we get:
m = 3.94 × 10^4 J / (418 J/g)
m ≈ 94.02 g
Therefore, the mass of liquid water required to absorb 3.94 × 10^4 Joules of heat energy on boiling is approximately 94.02 grams.