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:
8/11
Step-by-step explanation:
Only 3 of the nickels are neither dimes nor Canadian. The other 8 of 11 coins are dimes or Canadian. The probability of choosing one of them at random is 8/11.
Answer:
The F-statistic used to test the hypothesis that the miles per gallon for each fuel are the same is 4.07.
Step-by-step explanation:
There are four treatments in the data given, i.e. k = 4.
Total number of observations, n = 12.
Note: degrees of freedom is denoted as df.
For treatment, the degrees of freedom = k-1 = 4-1 =3 df.
The total degrees of freedom = n-1 = 12-1 = 11 df.
The error in degrees of freedom = df (total) - df(treatment)
The error in degrees of freedom = 11 - 3 = 8 df
At α = 0.05 level,from the F table, the F-statistic with (3 , 8)df is 4.07.
Therefore, the F-statistic used to test the hypothesis that the miles per gallon for each fuel are the same is 4.07.
I’ve been trying to figure out,but I can’t get it
