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:
Kindly check explanation
Step-by-step explanation:
Given the following :
Starting population = 4000
Addition per month = 170
decline on population per month = 70
Increase rate in population per month (dt) :
Starting population = 4000
Number of births per month = 170
However, the population declines by 70 individuals each month
Hence,
Number of births - number of deaths(d) = 70
170 - d = - 70 ( decline?
170 + 70 = d
240 = d
d = number of deaths
Per capita death :
Total number of deaths per. Month / starting population
= 240 / 4000
= 0.06
Percent increase
find increase first
10500 to 11300
11300-10500=800
so
percent increase
change/original
origianal=10500
change=800
800/10500=8/105=0.0761
percent means parts out of 100
0.0761/1 times 100/100=7.61/100=7.61%
rond 7.61% to tenth or to 7.6%
7.6%
Answer:
42 i think
Step-by-step explanation:
this is sssniperwolf im on to talk to fans
sorry if its incorrect :3