The amount of heat required to convert H₂O to steam is : 382.62 kJ
<u>Given data :</u>
Mass of liquid water ( m ) = 150 g
Temperature of liquid water = 43.5°C
Temperature of steam = 130°C
<h3 /><h3>Determine the amount of heat required </h3>
The amount of heat required = ∑ q1 + q2 + q3 ----- ( 1 )
where ;
q1 = heat required to change Temperature of water from 43.5°C to 100°C . q2 = heat required to change liquid water at 100°C to steam at 100°C
q3 = heat required to change temperature of steam at 100°C to 130°C
M* S
*ΔT
= 150 * 4.18 * ( 100 - 43.5 )
= 35425.5 J
moles * ΔHvap
= (150 / 18 )* 40.67 * 1000
= 338916.67 J
M * S
* ΔT
= 150 * 1.84 * ( 130 -100 )
= 8280 J
Back to equation ( 1 )
Amount of heat required = 35425.5 + 338916.67 + 8280 = 382622.17 J
≈ 382.62 kJ
Hence we can conclude that The amount of heat required to convert H₂O to steam is : 382.62 kJ.
Learn more about Specific heat of water : brainly.com/question/16559442
TLDR: The kinetic energy is determined to be zero.
Kinetic energy is energy of motion; when an object is moving (i.e. it has speed or velocity), it has some amount of kinetic energy. The equation itself looks like so:
KE=1/2(m)(v)^2,
where "m" represents the mass of the object and "v" represents the objects speed or velocity. In this example, the ball has stopped, meaning it has no speed/velocity. This means that the final kinetic energy is determined to be zero or none, due to the lack of motion. Mathematically, you can see this by substituting "0" in for "v" (the ball is stopped):
KE=1/2(m)(v)^2
KE=1/2(m)(0)^2
KE=1/2(m)*0
KE=1/2*0
KE=0 J,
or zero kinetic energy.
Hope this helps! :)