Answer:
T₂ = 36.36°C
Explanation:
The complete question is:
<u>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.</u>
<u>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? </u>
Answer: 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)
<u>T₂ = 36.36°C</u>
