If 150.0 g of copper at 100.0 C and 125.0g of liquid water at 35.0 C are mixed in an insulated container, what is the final temperature of the mixture? Ccu= .39 J/g C Cwater= 4.184 J/g C
Did i do this correct:
150.0 x 100.0 x .39 = 250 J
125.0 x 35.0 x 4.184 = 18300 J
18300 + 250= 18550 J
I used sig fig.
The first multiplication should be higher
i did
100j/.39x150.0g=1.7c use sig fig so its 2
then you subtract from the H2O the temp. change
To calculate the final temperature of the mixture, you need to apply the principle of conservation of energy. The heat lost by the copper will be gained by the water until thermal equilibrium is reached. The heat gained or lost is given by the equation:
q = m * c * ΔT
Where:
q is the heat gained or lost
m is the mass
c is the specific heat capacity
ΔT is the change in temperature
First, calculate the heat lost by the copper and the heat gained by the water:
For copper:
q_copper = (mass_copper) * (specific_heat_copper) * (T_final - T_initial)
q_copper = 150.0 g * 0.39 J/g°C * (T_final - 100.0°C)
For water:
q_water = (mass_water) * (specific_heat_water) * (T_final - T_initial)
q_water = 125.0 g * 4.184 J/g°C * (T_final - 35.0°C)
Since energy is conserved, the heat lost by the copper should be equal to the heat gained by the water:
q_copper = q_water
150.0 g * 0.39 J/g°C * (T_final - 100.0°C) = 125.0 g * 4.184 J/g°C * (T_final - 35.0°C)
Now, solve for T_final:
(150.0 g * 0.39 J/g°C * T_final) - (150.0 g * 0.39 J/g°C * 100.0°C) = (125.0 g * 4.184 J/g°C * T_final) - (125.0 g * 4.184 J/g°C * 35.0°C)
Simplify the equation:
(195.0 T_final) - 5850.0 = (522.5 T_final) - 18207.5
Combine like terms:
(195.0 - 522.5) T_final = (5850.0 - 18207.5)
Now, solve for T_final:
(-327.5) T_final = -12357.5
Divide both sides of the equation by -327.5:
T_final = (-12357.5) / (-327.5)
T_final ≈ 37.76°C
Therefore, the final temperature of the mixture is approximately 37.76°C.