Water evaporates at 100⁰C
So change in temperature = 100-20 = 80⁰C
Amount of water to be evaporated = 1 liter = 1L*1kg/liter = 1 kg
Specific heat of water is 1 calorie/gram ⁰C = 4.186 joule/gram =4186 J/kg
So heat required E = mcΔT = 1 * 4186 *80= 334880 J =334.88 kJ
So amount of heat require to evaporate water = 334.88 kJ
<h3><u>Answer</u>;</h3>
$347.22
<h3><u>Explanation</u>;</h3>
Principal = $14,200
Rate = 8.5%
Time = 105 days = 105/365
Interest = Principal x Rate x Time
Interest = 14,200 x 0.085 x 105/365
Interest = 347.219
= $347.22
Explanation:
We'll need two equations.
v² = v₀² + 2a(x - x₀)
where v is the final velocity, v₀ is the initial velocity, a is the acceleration, x is the final position, and x₀ is the initial position.
x = x₀ + ½ (v + v₀)t
where t is time.
Given:
v = 47.5 m/s
v₀ = 34.3 m/s
x - x₀ = 40100 m
Find: a and t
(47.5)² = (34.3)² + 2a(40100)
a = 0.0135 m/s²
40100 = ½ (47.5 + 34.3)t
t = 980 s
