A solid sphere of uniform density starts from rest and rolls without slipping down an inclined plane with angle θ = 30o. The sphere has mass M = 8 kg and radius R = 0.19 m . The coefficient of static friction between the sphere and the plane is μ = 0.64. What is the magnitude of the frictional force on the sphere?
Ff =
Solution: From the question we have
A pot with a steel bottom 8.50 mm thick rests on a hot stove.
The area of the bottom of the pot is 0.155 m2
The water inside the pot is at 100 °C, and 0.370 kg are evaporated every 2.90 min
To Find: the temperature of the lower surface of the pot, which is in contact with the stove.
The heat flow needed to evaporate the given amount of water is
H = ΔQ/Δt = Δh_vap·Δm/Δt
Latent heat of vaporization of water at 100°C is [1]
Δh_vap = 2257kJ/kg
hence:
H = 2257kJ/kg · 0.370kg / (3·60s)
= 4.799 kJ/s
= 4799.36 W
Assuming one-dimensional heat flow through the bottom of the pot, Fourier's law of heat conduction takes the form [2]:
H = k·A·ΔT/Δx = k·A·(T - T0)/Δx
Supposed the pot is made of stainless steel, its bottom has a thermal conductivity of:
k = 15W/m°C
Solve equation above for T:
H = k·A·(T - T0)/Δx
=>
T = T0 + H·Δx / (k·A)
= 100°C + 4799.36W·0.01m/(15W/m°C·0.01m²)
= 147.99°C answe