Answer:
W= 4.89 KJ
Explanation:
Lets take
temperature of hot water T₁ = 100⁰C
T₁ = 373 K
Temperature of cold ice T₂= 0⁰C
T₂ = 273 K
The latent heat of ice LH= 334 KJ
The heat rejected by the engine Q= m .LH
Q₂= 0.04 x 334
Q₂= 13.36 KJ
Heat gain by engine = Q₁
For Carnot engine
Q₁ = 18.25 KJ
The work W= Q₁ - Q₂
W= 18.25 - 13.36 KJ
W= 4.89 KJ
Answer:
The minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Explanation:
We know by equation of motion that,
Where, v= final velocity m/sec
u=initial velocity m/sec
a=Acceleration m/
s= Distance traveled before stop m
Case 1
u= 13 m/sec, v=0, s= 57.46 m, a=?
a = -1.47 m/ (a is negative since final velocity is less then initial velocity)
Case 2
u=29 m/sec, v=0, s= ?, a=-1.47 m/ (since same friction force is applied)
s = 285.94 m
Hence the minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m