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
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
woah!
Thank you Elena!
To find the equilibrium temperature of the mixture, we can use the principle of conservation of energy. The total heat gained by the water, copper, and aluminum must be equal to the total heat lost.
Let's calculate the heat gained by the water:
Q_water = mass_water x specific_heat_water x ΔT_water
We are given:
mass_water = 5.0 x 10^2 g
specific_heat_water = value from the table (let's assume it's 4.18 J/g∙K)
ΔT_water = (equilibrium temperature - initial temperature of water)
Next, let's calculate the heat lost by the copper and aluminum:
Q_losing = mass_copper x specific_heat_copper x ΔT_copper + mass_aluminum x specific_heat_aluminum x ΔT_aluminum
We are given:
mass_copper = 1.0 x 10^2 g
specific_heat_copper = 0.215 J/g∙K (from the table)
ΔT_copper = (equilibrium temperature - initial temperature of copper)
mass_aluminum = 2.0 x 10^2 g
specific_heat_aluminum = 0.900 J/g∙K (from the table)
ΔT_aluminum = (equilibrium temperature - initial temperature of aluminum)
The total heat gained by the water (Q_water) is equal to the total heat lost by the copper and aluminum (Q_losing) at equilibrium.
We can set up the equation as follows:
mass_water x specific_heat_water x ΔT_water = mass_copper x specific_heat_copper x ΔT_copper + mass_aluminum x specific_heat_aluminum x ΔT_aluminum
Substituting the known values:
5.0 x 10^2 g x 4.18 J/g∙K x (equilibrium temperature - 280 K) = 1.0 x 10^2 g x 0.215 J/g∙K x (equilibrium temperature - 460 K) + 2.0 x 10^2 g x 0.900 J/g∙K x (equilibrium temperature - 2.2 x 10^2 K)
Now, let's solve for the equilibrium temperature (tf):
Multiply out and rearrange the equation to isolate tf:
2090 x (tf - 280) = 21.5 x (tf - 460) + 180 x (tf - 220)
2090tf - 581,200 = 21.5tf - 9,815 + 180tf - 39,600
Combine like terms and move constants to one side:
2090tf - 21.5tf - 180tf = 581,200 - 9,815 - 39,600
1888.5tf = 531,785
Divide both sides by 1888.5:
tf = 531,785 / 1888.5
tf ≈ 281.4 K
Therefore, the equilibrium temperature of the mixture is approximately 281.4 K.