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
C. Region D
Step-by-step explanation:
Just plug the equations into desmos the calculator will graph it and show you with colors where the solution is.
Answer:
Use this to solve 3
Step-by-step explanation:
Area = 1/2 b.c sin A
=1/2(13.1)(8.7)sin29. 17
<span>Successive discounts of 20% and 10% are taken on an item priced at $16.
=> Let's find out how much is the discount in all.
=> 16 dollars * .20 = 3.2 dollars
=> 16 - 3.2 = 12.8 dollars.
then another 10% discount,
=> 12.8 * .10 = 1.28 dollars
=> 12.8 - 1.28 = 11.52 dollars is not the price minus the discounts,</span>