Answer:
7479 cal.
31262.2 joules
Explanation:
This is a calorimetry problem where water in its three states changes from ice to vapor.
We must use, the calorimetry formula and the formula for latent heat.
Q = m . C . ΔT
Q = Clat . m
First of all, let's determine the heat for ice, before it melts.
10 g . 0.5 cal/g°C ( 0° - (-20°C) = 100 cal
Now, the ice has melted.
Q = Clat heat of fusion . 10 g
Q = 79.7 cal/g . 10 g → 797 cal
We have water at 0°, so this water has to receive heat until it becomes vapor. Let's determine that heat.
Q = m . C . ΔT
Q = 10 g . 1 cal/g°C (100°C - 0°C) → 1000 cal
Water is ready now, to become vapor so let's determine the heat.
Q = Clat heat of vaporization . m
Q = 539.4 cal/g . 10 g → 5394 cal
Finally we have vapor water, so let's determine the heat gained when this vapor changes the T° from 100°C to 120°
Q = m . C . ΔT
Q = 10 g . 0.470 cal/g°C . (120°C - 100°C) → 94 cal
Now, we have to sum all the heat that was added in all the process.
100 cal + 797 cal + 1000 cal + 5394 cal + 94 cal =7479 cal.
We can convert this unit to joules, which is more acceptable for energy terms.
1 cal is 4.18 Joules.
Then, 7479 cal are (7479 . 4.18) = 31262.2 joules
- Ideal gas constant, in kilopascal-cubic meters per kilomole-Kelvin.
- Temperature, in Kelvin
- Pressure, in kilopascals
