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:
x = 5, x = 15; The zeros represent the number of photos printed to produce a maximum profit.
Step-by-step explanation:
F(x) = −x²+ 20x − 75
Let's f(x)= 0
0= -x²+20x-75
0= x²-20x+75
0=x²-5x-15x+75
0= x(x-5)-15(x-5)
0= (x-5)(x-15)
The zeros are
X= 5 or x = 15
And the xeros represent a number of photos printed to produce maximum profit