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
Assuming the velocity is as written,
a = 3(2) = 6
If you meant 3(2t+1)⁰.⁴, then
a = 2.4/(2t+1)⁰.⁶
as usual,
s = 25 + 1/2 at²
= 25 + 3t²
or
= 25 + 1.2(2t+1)¹.⁴
The statment is correct but if you have to work out the value of x you need a total amount of the purchase without that or atleast some sort of value it is not possible to work out
The rule of 72 is an approximate estimate of the time it takes to double an investment, and depends only on the interest rate. So amount of deposit does not change the estimate. All three accounts will take the same time to double.
If the accounts are all deposited on the same day with the same interest rate and same compounding period, they all double at the same time, whether using the rule of 72 or the actual time.