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:
10 to the 4th power
Step-by-step explanation:
the decimil point was moved four spaces to the right resutiong to the 4th power.
Answer:
250 ft
Step-by-step explanation:
circumference of circle 'C' is defined as
C = 2πr
250π = 2πr
r = 250π/2π
r= 125
d = 2r = 2×125 = 250
.
diameter of the given circle is 250 ft
Answer:
y – 1 = y minus 1 equals StartFraction one-half EndFraction left-parenthesis x minus 5 right-parenthesis.(x –5)
Step-by-step explanation:
we know that
The equation of the line into point slope form is equal to

we have


substitute

Answer:
Im not sure about 20 but 22 is 48
Step-by-step explanation:
4×96 =384
384÷8 = 48 orange per classroom