Answer:
2.36 x 10^6 J
Explanation:
Tc = 0°C = 273 K
TH = 22.5°C = 295.5 K
Qc = heat used to melt the ice
mass of ice, m = 85.7 Kg
Latent heat of fusion, L = 3.34 x 10^5 J/kg
Let Energy supplied is E which is equal to the work done
Qc = m x L = 85.7 x 3.34 x 10^5 = 286.24 x 10^5 J
Use the Carnot's equation
QH = 309.8 x 10^5 J
W = QH - Qc
W = (309.8 - 286.24) x 10^5
W = 23.56 x 10^5 J
W = 2.36 x 10^6 J
Thus, the energy supplied is 2.36 x 10^6 J.
Answer:
w = 5832.372 Joules
Explanation:
Mass of water, m = 20 kg
The water was pulled up to a height of 35 meters, i.e. h = 35 m
It takes 14 minutes to pull up the water through the height, 35 m
speed = distance/ time = 35/14 = 2.5 m/min
The bucket's height, y = speed * time = 2.5t meters
6 kg of water drips out of the bucket throughout the 14 minutes
The rate at which the water drips drips out = (6/14) = 0.4286 kg/min
Mass of water that drips out in time, t = 0.4286t kg
The mass of water remaining = (20 - 0.4286t) kg
Change in Workdone, Δw = mgΔy
Δy = 2.5 Δt
Δw = mg * 2.5 Δt
dw = (20 - 0.4286t)g2.5 dt
integrating both sides
dw = (50g - 1.07gt)dt
where b = 0, a = 14
w = 50gt - 1.07g(t²)/2 g = 9.8 m/s²
w = 490t - 5.243t²
w = (490*14 - 5.243*14²) - (490*0 - 5.243*0²)
w = 6860 - 1027.628
w = 5832.372 Joules