If 5.40 kcal of heat is added to 1.00 kg of water at 100⁰C, how much steam at 100⁰C is produced? Show all calculations leading to an answer.
Please help?
massSteam*.540kcal/kg =5.4kcal
looks like massSteam=10kg.
check the heat of vaporization, I haven't used Hv in kcal/kg in 50 years. My memory fades.
To solve this problem, we need to use the concept of heat transfer and specific heat capacity.
First, let's find the heat absorbed by water using the formula:
Q = m * c * ΔT
where Q denotes the heat absorbed, m is the mass of the substance (water in this case), c is the specific heat capacity, and ΔT is the change in temperature.
Given:
Q = 5.40 kcal
m = 1.00 kg
ΔT = (final temperature - initial temperature)
The specific heat capacity of water is approximately 4.18 J/g°C (or 1.00 cal/g°C). However, we need to convert kcal to cal and grams to kilograms.
1 kcal = 1000 cal
1 kg = 1000 g
Converting the given values:
Q = 5.40 kcal * 1000 cal/kcal = 5400 cal
m = 1.00 kg * 1000 g/kg = 1000 g
Next, we need to determine the change in temperature.
At 100⁰C, water is already at its boiling point. So, the change in temperature is 100⁰C – 100⁰C = 0⁰C.
Plugging in these values into the formula:
5400 cal = 1000 g * 4.18 J/g°C * 0°C
Simplifying the equation:
5400 cal = 0 J
Since the equation does not balance, it means the water absorbs all the heat without any temperature change. Therefore, no steam is produced.
Hence, the answer is that no steam is produced when 5.40 kcal of heat is added to 1.00 kg of water at 100⁰C.