A 1.0 102 g piece of copper at a temperature of 460 K and a 2.0 102 g piece of aluminum at a temperature of 2.2 102 K are dropped into an insulated bucket containing 5.0 102 g of water at 280 K. What is the equilibrium temperature of the mixture? (Use the specific heats found in this table.)
I have 252.2tf=57692
tf=228 K
Based off of the table the specific heats for Aluminum and Copper are. Al-0.900 and Copper-0.215
6 answers
And the answer that I have is wrong
I'm sorry for Aluminum its 0.215 and for Copper its 0.0922
The grams for Aluminum are 2.0e2 and the temperature for Aluminum is 2.2e2. The grams for water is 5.0e2.
Cu: m₁=0.1 kg, c₁=385 J/kg•K, T₁=460 K,
Al: m₂=0.2 kg, c₂=930 J/kg•K, T₂=220 K,
Water: m₃=0.5 kg, c₃=4183 J/kg•K, T₃280 K.
T=?
m₁c₁(T₁-T) = m₂c₂(T-T₂) +m₃c₃(T-T₃)
T= (m₁c₁T₁ + m₂c₂T₂ + m₃c₃T₃)/(m₁c₁ + m₂c₂ + m₃c₃)=
...=278.1 K
Al: m₂=0.2 kg, c₂=930 J/kg•K, T₂=220 K,
Water: m₃=0.5 kg, c₃=4183 J/kg•K, T₃280 K.
T=?
m₁c₁(T₁-T) = m₂c₂(T-T₂) +m₃c₃(T-T₃)
T= (m₁c₁T₁ + m₂c₂T₂ + m₃c₃T₃)/(m₁c₁ + m₂c₂ + m₃c₃)=
...=278.1 K
woah!
Thank you Elena!
1 answer