Answer:
The amount of heat required to raise the temperature of liquid water is 9605 kilo joule .
Step-by-step explanation:
Given as :
The mass of liquid water = 50 g
The initial temperature =
= 15°c
The final temperature =
= 100°c
The latent heat of vaporization of water = 2260.0 J/g
Let The amount of heat required to raise temperature = Q Joule
Now, From method
Heat = mass × latent heat × change in temperature
Or, Q = m × s × ΔT
or, Q = m × s × (
-
)
So, Q = 50 g × 2260.0 J/g × ( 100°c - 15°c )
Or, Q = 50 g × 2260.0 J/g × 85°c
∴ Q = 9,605,000 joule
Or, Q = 9,605 × 10³ joule
Or, Q = 9605 kilo joule
Hence The amount of heat required to raise the temperature of liquid water is 9605 kilo joule . Answer
Answer:
f(2) = 48
Step-by-step explanation:
f(2) = 3 * 4^2
f(2) = 3 * 16
f(2) = 48
Answer: f(2) = 48
Answer:
x=10
Step-by-step explanation:
<AOC =90°
and <AOB + <BOC = <AOC
3x+46+14=90°
3x+60=90°
3x = 90°-60°
3x=30°
X=30÷3
X= 10°
Answer: x < -26
Step-by-Step Explanation:
87-3x>165
-3x> (165-85)
-3x>80
80/-3 = -26.666
CHECK:
87-3*(-26)=165
x < -26
<em>Hope this helps!!!</em>