A 542-g piece of copper tubing is heated to 89.5 degrees Celcius and placed in an insulated vessel containing 159 g of water at 22.8 degrees Celcius. Assuming no loss of water and a heat capacity for the vessel of 10.0 J/K, what is the final temperature of the system (c=0.387 J/g*K)?
[mass copper x specific heat Cu x (Tfinal-Tinitial)] + [mass water x specific heat water x (Tfinal-Tinitial)] + [heat capacity vessel x 10 J/K] = 0
Watch the units.
To find the final temperature of the system, we need to use the principle of conservation of energy.
The heat lost by the copper tubing will be equal to the heat gained by the water and the vessel.
The formula for heat transfer is:
Q = m * c * ΔT
Where:
Q = heat transfer (in joules)
m = mass of the substance (in grams)
c = specific heat capacity (in J/g*K)
ΔT = change in temperature (in degrees Celsius)
Let's calculate the heat transfer for the copper tubing:
Q_tubing = m_tubing * c_tubing * ΔT_tubing
Q_tubing = 542 g * 0.387 J/g*K * (T_final - 89.5°C)
Next, let's calculate the heat gained by the water:
Q_water = m_water * c_water * ΔT_water
Q_water = 159 g * 4.18 J/g*K * (T_final - 22.8°C)
Finally, let's calculate the total heat transfer and set it to zero since there is no net heat transfer from the system:
Q_total = Q_tubing + Q_water + Q_vessel
Since Q_total = 0:
0 = Q_tubing + Q_water + Q_vessel
Now we can substitute the values into the equation and solve for T_final:
0 = (542 g * 0.387 J/g*K * (T_final - 89.5°C)) + (159 g * 4.18 J/g*K * (T_final - 22.8°C)) + (10.0 J/K * (T_final - 22.8°C))
Simplify and solve the equation to find the final temperature (T_final) of the system.