paula charges $12 to shovel a driveway she shoveled 6 driveaways on saturday and then some more on sunday she made $144 for the
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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
For this case what we should do is take into account the following conversion:
We then have the following number:
Applying the given conversion we have:
By doing the corresponding calculation we have that the standard form is given by:
Answer:
4540 million in standard form is:
We then have the following number:
Applying the given conversion we have:
By doing the corresponding calculation we have that the standard form is given by:
Answer:
4540 million in standard form is: