2m - nx = x + 4 with description
1 answer:
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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
Answer:
x = 7
Step-by-step explanation:
Given:
∠DEF = 117°
∠DEG = (12x + 1)°
∠GEF = (5x - 3)°
Find:
value of x
Computation:
∠DEF = ∠DEG + ∠GEF
117° = (12x + 1)° + (5x - 3)°
117° = 17 x - 2
x = 7
