A hot lump of 33.2 g of copper at an initial temperature of 70.5 °C is placed in 50.0 mL of H2O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the copper and water given that the specific heat of copper is 0.385 J/(g·°C)? Assume no heat is lost to surroundings.
[mc∆T]Cu + [mc∆T]HOH = 0
[mc(Tf - Ti)]Cu + [mc(Tf - Ti)]HOH = 0
All data is given in problem except T-final. Solve for Tf.
m(Cu) = 33.2 gm
m(HOH)= 50.0 gm
Ti(Cu) = 70.5 C
Ti(HOH)= 25.0 C
Tf = ???
c(Cu) = 0.385 J/gC
c(HOH)= 4.184 J/gC
Your answer should be between the temp extremes 25C & 70.5C.
To solve this problem, we need to use the principle of energy conservation, which states that the heat gained by one object equals the heat lost by another object in thermal contact until thermal equilibrium is reached.
The equation to calculate the heat (q) gained or lost by an object is:
q = m * c * ΔT
Where:
- q represents heat
- m is the mass of the object
- c is the specific heat of the substance
- ΔT is the change in temperature
In this case, we have two objects: the copper and the water. Let's calculate the heat gained or lost by each object:
1. Copper:
- Mass (m) = 33.2 g
- Specific heat (c) = 0.385 J/(g·°C)
- Initial temperature (T1) = 70.5 °C
- Final temperature (T2) = ?
- ΔT = T2 - T1
q₁ = m * c * ΔT
2. Water:
- Mass (m) = 50.0 g
- Specific heat (c) = 4.184 J/(g·°C) (specific heat of water)
- Initial temperature (T1) = 25.0 °C
- Final temperature (T2) = ?
- ΔT = T2 - T1
q₂ = m * c * ΔT
Since there is no heat lost to the surroundings, the heat gained by the copper will be equal to the heat lost by the water (q₁ = -q₂).
Setting up the equation and solving for T2:
m₁ * c₁ * ΔT₁ = -m₂ * c₂ * ΔT₂
m₁ * c₁ * (T₂ - T₁) = -m₂ * c₂ * (T₂ - T₁)
(33.2 g) * (0.385 J/(g·°C)) * (T₂ - 70.5 °C) = -(50.0 mL) * (4.184 J/(g·°C)) * (T₂ - 25.0 °C)
Now, let's solve for T₂:
(33.2 g) * (0.385 J/(g·°C)) * T₂ - (33.2 g) * (0.385 J/(g·°C)) * 70.5 °C = -(50.0 mL) * (4.184 J/(g·°C)) * T₂ + (50.0 mL) * (4.184 J/(g·°C)) * 25.0 °C
(12.782 g·°C) * T₂ - (8.533 g·°C) = -(209.2 J·°C) * T₂ + (522.4 J·°C)
(12.782 g·°C + 209.2 J·°C) * T₂ = (8.533 g·°C + 522.4 J·°C)
(222.982 g·°C) * T₂ = (530.933 g·°C)
T₂ = (530.933 g·°C) / (222.982 g·°C)
T₂ ≈ 2.378 °C
Therefore, the final temperature of the copper and water when they reach thermal equilibrium is approximately 2.378 °C.