Answer:
The heat needed to boil 1 gallon of water is 81,490.62 Joules.
Explanation:
Where:
Q = heat absorbed or heat lost
c = specific heat of substance
m = Mass of the substance
ΔT = change in temperature of the substance
We have :
Volume of water = V = 1 gal = 4546.09 mL
Density of water , d= 1 g/mL
mass of water = m = d × V = 1g/mL × 4546.09 mL = 4546.09 g
Specific heat of water = c = 1 Cal/g°C
ΔT = 100°C - 25°C = 75 °C
9 (boiling pint of water is 100°C)
Heat absorbed by the water to make it boil:
1 calorie = 4.184 J
The heat needed to boil 1 gallon of water is 81,490.62 Joules.
Answer: It changed in identity and properties.
Explanation:
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Answer:
Check explanation
Explanation:
From the question, the parameters given are 64.7g of benzene,C6H6; a starting temperature of 41.9°C and bringing it to 33.2°C.
Molar mass of benzene,C6H6= 78.11236 g/mol.
Things to know: heat capacity of benzene, C6H6= 1.63 J/g.K, the heat of fusion = 9.87 kj/mol.
STEP ONE(1): ENERGY USED IN MELTING BENZENE SOLID.
Using the formula below;
Energy used in melting the solid(in JOULES) = (mass of benzene/molar mass of benzene) × heat of Fusion.
=(64.7 g of C6H6/ 78.11236(g per mol) of C6H6) × 9.87 kJ per mol.
= 8.175 J.
= 0.008175 kJ.
STEP TWO (2): ENERGY OF HEATING THE LIQUID.
It can be calculated from the formula below;
Energy= heat capacity (J/g.K) × mass of benzene× (∆T).
= 1.63 J/g.K × 64.7 × (41.9-33.2).
= 917.5J.
= 0.9175 kJ.
Energy required to boil benzene= Energy required to melt the bezene + energy required for boiling.
= 0.008175+ 0.9175.
= 0.93kJ
Approximately, 1 kJ
Answer:
the reactants are 2h2 and 02. the products are 2h20
have a great day my friend ;)
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