To solve this problem, we should recall the law of
conservation of energy. That is, the heat lost by the aluminium must be equal
to the heat gained by the cold water. This is expressed in change in enthalpies
therefore:
- ΔH aluminium = ΔH water
where ΔH = m Cp (T2 – T1)
The negative sign simply means heat is lost. Therefore we
calculate for the mass of water (m):
- 0.5 (900) (20 – 200) = m (4186) (20 – 0)
m = 0.9675 kg
Using same mass of water and initial temperature, the final
temperature T of a 1.0 kg aluminium block is:
- 1 (900) (T – 200) = 0.9675 (4186) (T – 0)
- 900 T + 180,000 = 4050 T
4950 T = 180,000
T = 36.36°C
The final temperature of the water and block is 36.36°C
Answer:
The “terminal speed” of the ball bearing is 5.609 m/s
Explanation:
Radius of the steel ball R = 2.40 mm
Viscosity of honey η = 6.0 Pa/s
While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)
Substitute the given values to find "terminal speed"
The “terminal speed” of the ball bearing is 5.609 m/s
The object does not move.
It means that you consider the elements as a list organized by atomic number, the property is seen to repeat over and over as you move through that list.
