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
Answer:
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Answer: the density of the rock
55.91g/ml
Explanation:
The density of the rock after it had being placed in the cylinder of water so the calculation should look like this:
Volume of water substract the mass of the rock:
And that is 142.5 ml - 86.59g =
Answer 55.91 g/ml
So the density of the rock is 55.91g/ml
<u>S</u><u>o</u><u>l</u><u>u</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u><u>-</u><u>1</u>

- E denotes to Energy
- h denotes to planks constant
- v denotes to frequency.
<u>S</u><u>o</u><u>l</u><u>u</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u><u>-</u><u>2</u>





Now



There are 2.3 moles (because of significant figures)