A0.5 kg block of aluminum (caluminum=900j/kg⋅∘c) is heated to 200∘c. the block is then quickly placed in an insulated tub of cold water at 0∘c (cwater=4186j/kg⋅∘c) and sealed. at equilibrium, the temperature of the water and block are measured to be 20∘c. part a if the original experiment is repeated with a 1.0 kg aluminum block, what is the final temperature of the water and block?

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

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

1/0.14nm = 10^10/1.4 a lot