If the original experiment is repeated with a 1.0 kg aluminum block, what is the final temperature of the water and block

T₂ = 36.36°C

Explanation:

The complete question is:

A 0.5 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 0C (Cwater=4186J/kg*C) and sealed. At equilibrium, the temperature of the water and block are measured to be 20C.

A)     If the original experiment is repeated with a 1.0 kg {kg} aluminum block, what is the final temperature of the water and block?

We will apply the law of conservation of energy here:

Heat Loss by Aluminum Block = Heat Gained by Water

m₁C₁ΔT₁ = m₂C₂ΔT₂

where,

m₁ = mass of aluminum block

C₁ = Specific Heat Capacity of Aluminum Block = 900 J/kg °C

ΔT₁ = Change in Temperature of Aluminum Block

m₂ = mass of water

C₂ = Specific Heat Capacity of Water = 4186 J/kg °C

ΔT₂ = Change in Temperature of water

Therefore, for initial given condition:

m₁ = 0.5 kg

ΔT₁ = 200°C - 20°C = 180°C

m₂ = ?

ΔT₂ = 20°C - 0°C = 20°C

Therefore,

(0.5 kg)(900 J/kg °C)(180°C) = m₂(4186 J/kg °C)(20°C)

m₂ = 81000 J/(83720 J/kg)

m₂ = 0.97 kg

Now, we have the mass of water. So, we can solve for the final equilibrium temperature if the mass of block is changed to 1 kg. So, for the new state:

m₁ = 1 kg

m₂ = 0.97 kg

ΔT₁ = 200°C - T₂

ΔT₂ = T₂ - 0°C

T₂ = Final Equilibrium Temperature = ?

Therefore,

(1 kg)(900 J/kg °C)(200°C - T₂) = (0.97 kg)(4186 J/kg °C)(T₂ - 0°C)

180000 J  - (900 J/°C)(T₂) = (4050 J/°C)(T₂)

180000 J = (900 J/°C)(T₂) + (4050 J/°C)(T₂)

T₂ = (180000 J)/(4950 J/°C)

T₂ = 36.36°C

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more