Water falls from a height of 50m. Calculate the rise in temperature of water when it strikes the bottom, g=10m/s2 specific heat capacity of water = 4200j/kg/C
Answers
Answered by
153
We can take that work done is equal
So,
mgs=mc∆t
therefore
∆t=gs/c
g=10,s=50,c=4200
answer ∆t=0.119 °C
So,
mgs=mc∆t
therefore
∆t=gs/c
g=10,s=50,c=4200
answer ∆t=0.119 °C
Answered by
35
Answer:
The temperature of water rises about 0.12 degree C when it strikes the bottom.
Explanation:
The water fallen gains temperature and strikes bottom with an increased temperature, calculated as follows –
Given that the height of fall h = 50 m
Acceleration g = 10 m/ s 2
Let the water falling has a mass of m.
Thereby, the energy of water E is given by –
E = m g h
E = m x 10 x 50
E = 500 m
Now this is the amount of energy which is converted to the heat causing the rise in temperature, so we have –
E = δQ
E =δQ
E = m s δ T
Given s is the specific heat capacity of water as s = 4200 j / kg / C
E = m s δT
500 m = m 4200 δ T
δ T = 500 / 4200
δ T = 5 /42
δT = 0.12 degree C.
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