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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.
<h3>Answer:</h3>
0.125 mol Ca
<h3>General Formulas and Concepts:</h3><u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<u>Step 1: Define</u>
5.00 g Ca
<u>Step 2: Identify Conversions</u>
Molar mass of Ca - 40.08 g/mol
<u>Step 3: Convert</u>
- Set up:
- Multiply:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.12475 mol Ca ≈ 0.125 mol Ca