A 100.0 gram sample of iron at 80.0 degrees Celsius was placed in 150 grams of water at 20.5 degrees Celsius. What was the final temperature when thermal equilibrium was reached?
To find the final temperature when thermal equilibrium is reached, we can use the principle of conservation of energy.
The heat gained by the water is equal to the heat lost by the iron when they reach thermal equilibrium.
The heat gained or lost can be calculated using the formula:
Q = m * c * ΔT
Where:
Q is the heat gained or lost
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in J/g°C)
ΔT is the change in temperature
First, let's calculate the heat gained by the water when it reaches thermal equilibrium.
Mass of water (m₁) = 150 grams
Initial temperature of water (T₁) = 20.5°C
Final temperature of water (T₂) = ?
Heat gained by water = m₁ * c_water * ΔT_water
The specific heat capacity of water (c_water) is approximately 4.184 J/g°C.
ΔT_water = T₂ - T₁
Now let's calculate the heat lost by the iron when it reaches thermal equilibrium.
Mass of iron (m₂) = 100 grams
Initial temperature of iron (T₃) = 80.0°C
Final temperature of iron (T₄) = ?
Heat lost by iron = m₂ * c_iron * ΔT_iron
The specific heat capacity of iron (c_iron) is approximately 0.45 J/g°C.
ΔT_iron = T₃ - T₄
Since the heat gained by the water is equal to the heat lost by the iron, we can set up the equation:
m₁ * c_water * ΔT_water = m₂ * c_iron * ΔT_iron
Now, we can substitute the values and solve for the final temperature (T₂ and T₄):
150 * 4.184 * (T₂ - 20.5) = 100 * 0.45 * (80.0 - T₄)
Simplify the equation:
627.6T₂ - 12767.8 = 36(80 - T₄)
627.6T₂ - 12767.8 = 2880 - 36T₄
Combine like terms:
663.6T₂ + 36T₄ = 15547.8
We have two equations with two unknowns. Solving this system of equations will give us the final temperatures of both substances.