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:
288 gallons ÷ 6 miles = 48 miles per gallon
48 miles per gallon • 7 miles = 366 miles
Step-by-step explanation:
1748 people are expected to stop at the gift shop out of 2300.
Step-by-step explanation:
Given,
Number of random sampled people = 350
Number of people who stopped at shop = 266
Percentage = 
Percentage = 
Therefore,
76% people would stop at the gift stop.
Number of people = 2300
Expected number = 76% of total
Expected number = 
Expected number = 1748
1748 people are expected to stop at the gift shop out of 2300.
The answer is positive 25 and negative 25