The heat of fusion of ice is 79.7 cal/g. The heat of vaporization of water is 540 cal/g. Report the answer using the correct number of significant figures! 1. How much energy is required to melt 100.0 grams of ice? Mass of ice = 100.0g Heat of fusion of ice: 100,0g|79.7 call-(1970 cal) 2. How much energy is required to vaporize 234.5 g of water? 234.5g | 540cal |= :79.7 cal/g Answer: 7970 cal Mass of 234.59 Heat of vaporization & 540 callg 12 6630, 1.2663X105 (1.3 x 10³ cal Answer: 1.3 x 105 cal 3. If 30.6 calories are required to vaporize 25 g of a substance, what is the heat of vaporization of that substance? mass of substance: 25g Heat vaporization: 30
Thermochemistry
Thermochemistry can be considered as a branch of thermodynamics that deals with the connections between warmth, work, and various types of energy, formed because of different synthetic and actual cycles. Thermochemistry describes the energy changes that occur as a result of reactions or chemical changes in a substance.
Exergonic Reaction
The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.
![Calculations for Temperature and Phase Change Worksheet
The heat of fusion of ice is 79.7 cal/g.
The heat of vaporization of water is 540 cal/g.
Report the answer using the correct number of significant figures!
1. How much energy is required to melt 100.0 grams of ice? Mass of ice = 100.0g
Heat of fusion of ice:
100,0 g 79.7cal-1970 cal
2. How much energy is required to vaporize 234.5 g of water?
234.5g | 540 cal |-
30.6 cal
259
1.224
1.2 cal/g
1266301 1.2663X105
(1.3 x 10³ cal
Answer: 1.3 x 105 cal
3. If 30.6 calories are required to vaporize 25 g of a substance, what is the heat of vaporization of
that substance?
mass of substance: 25g
Heat
vaporization: 30.6 cal
-500.0g (1 cal) (1.1008) =
до
- 550 Ca1/4.1845
79.7cal/g
Answer: 7970 cal
Mass of 234.59
Heat of vaporization: 540 cal/g
Answer: 1.2 cal/g
4. How much energy is removed from 500.0 g of water when the temperature is lowered by 1.10 °C?
-F550 cal
mass m = 500.09
AT= 1.10°C
1000.09 1Cal (3x) = 3.00×10²cal
3%
Q==ssocal
Answer: -550. cal (or -2.30 x 10³ J)
23.01.2-2,3012×10³
-2.30× 10³5
Icg|
5. How much energy is required to raise the temperature of 1000.0 g of water from 23.00 °C to
26.00 °C?
m=1000.09
Q=S.M.AT
S=1 cal/goe Teinal - Tinitial
T₁ = 23.00°C
12=26.00°C
Answer: 3.00 x 103 cal (or 1.26 x 10¹J)
1.26×10 J
26.00-23.00=3%
8° 3.00x 10³ cal 4 1845 11.2.5 52x104
I cal
6. The specific heat of copper is (0.0924 cal/g°C), how much energy is required to raise the
temperature of 10.0 g of copper by 100.0 °C?
0.0924 cal
10.09
(357) (100.086)
92.4 call
mass of copper: 109
Specific heat copper: 0.0924cal/g °C
Answer: 92.4 cal](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd0ef70fe-2f8e-4de7-90a9-65ab0d73c4b0%2Fe1412926-7f6e-4ebb-9b43-9f05493d202d%2Fysrgoj8_processed.jpeg&w=3840&q=75)
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