A 36.0 g piece of copper at 130.0 C is placed in 50.0 g of water at 20.0 C. What is the temperature (in C) of the water once the system reaches equilibrium? The molar heat capacity of copper is 24.5 J/(mol·K). The specific heat capacity of water is 4.18 J/(g·K). The error interval is +/- 1C.
heat lost by Cu + heat gained by H2O 0
[mass Cu x specific heat Cu x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for Tfinal
0.9
To find the temperature of the water once the system reaches equilibrium, we can use the principle of conservation of energy. The energy lost by the copper is equal to the energy gained by the water. We can use the formula:
Q(copper) = -Q(water)
where Q represents heat, and the negative sign is because the copper loses heat while the water gains heat.
First, let's calculate the heat lost by the copper.
Q(copper) = mcΔT
where m is the mass of the copper, c is the specific heat capacity of copper, and ΔT is the change in temperature.
m(copper) = 36.0 g
c(copper) = 24.5 J/(mol·K)
Since we know the molar mass of copper is 63.55 g/mol, we can convert the mass of the copper to moles:
moles(copper) = m(copper) / molar mass(copper)
moles(copper) = 36.0 g / 63.55 g/mol
Next, we need to calculate the change in temperature of the copper. Since we know the initial temperature of the copper (130.0°C) and we need to find the final temperature, we can rewrite the formula as:
Q(copper) = mcΔT
Q(copper) = moles(copper) * c(copper) * ΔT(copper)
Similarly, let's calculate the heat gained by the water.
Q(water) = mcΔT
where m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
m(water) = 50.0 g
c(water) = 4.18 J/(g·K)
ΔT(water) = ? (to be determined)
Now we can set up the equation using the principle of conservation of energy:
moles(copper) * c(copper) * ΔT(copper) = m(water) * c(water) * ΔT(water)
Substituting the known values:
(36.0 g / 63.55 g/mol) * (24.5 J/(mol·K)) * (130.0°C - ΔT(copper)) = (50.0 g) * (4.18 J/(g·K)) * (ΔT(water) - 20.0°C)
Simplifying the equation:
(36.0 g / 63.55 g/mol) * (24.5 J/(mol·K)) * (130.0°C - ΔT(copper)) = (50.0 g) * (4.18 J/(g·K)) * (ΔT(water) - 20.0°C)
Solving for ΔT(water):
(36.0 g / 63.55 g/mol) * (24.5 J/(mol·K)) * (130.0°C - ΔT(copper)) = (50.0 g) * (4.18 J/(g·K)) * (ΔT(water)) - (50.0 g) * (4.18 J/(g·K)) * (20.0°C)
Now we can simplify the equation and solve for ΔT(water).