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:
Step-by-step explanation:
$27.63 / 2 = $13.82
$13.82 * 0.20 = $2.76
$13.82 + $2.76 = $16.58
To solve this we use trigonometric functions that would relate the hypotenuse y and the given values. For this case we use cosine function which is expressed as:
cosine theta = adjacent side / hypotenuse
cosine 52 = 35 / y
y = 35 / cos 52
y = 56.85
Answer:
Coordinates of other end point L = ( -19, 13 )
Explanation:
The mid point of the coordinates (a,b) and (c,d) is given by 
Here we have (a,b) = (1,-7) and
= (-9,3) , we need to find (c,d).

So coordinates of other end point L = ( -19, 13 )