A 32.5g cube of aluminum initially at 45.8 C is submerged into 105.3g water at 15.4 C. What is the final temp of both substances at thermal equilibrium. Assume the al and the H2O are thermally isolated from everything else)
Specific heat cap Al= .903
Specific heat cap H2O= 4.18
I understand this problem but In my math i get to:
(DeltaT) Al = -14.998 X (DeltaT) H2O
And im lost. Anyone able to explain the completion of the problem? Any help is greatly appreciated
Yes. I think it is tough to try to manipulate the algebra. It is much easier to substitute the numbers first and manipulate them.
[32.5 x 0.903 x (Tf-45.8)]+[105.3 x 4.18 x (Tf-15.4)] = 0
29.35Tf - 1344.1 + 440.15Tf - 6778.4 = 0
Check those numbers to make sure I didn't make an error on my calculator, then solve for Tf.
I'm having trouble once i get to
T(final)=-14.998 X T(final) + 14.998 X T(initial h2o) + T(initial Al)
Any suggestions?
You need to use Tfinal-Tinitial for delta T.
heat lost by Al + heat gained by water=0
[mass Al x specific heat Al x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for Tf.
Thank you so much DrBob!
I ended up with
469.5Tf = 8122.54
Tf = 17.3
Thanks again,
-Phil
To solve this problem, you need to use the principle of conservation of energy, which states that energy is conserved in a closed system. In this case, the energy gained by the aluminum is equal to the energy lost by the water.
First, you need to calculate the energy gained or lost by each substance. The energy gained or lost can be found using the formula:
Q = mcΔT
where Q is the energy gained or lost, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.
For the aluminum cube:
Q_al = m_al * c_al * ΔT_al
For the water:
Q_water = m_water * c_water * ΔT_water
Since the aluminum and water are in thermal equilibrium, the energy gained by the aluminum is equal to the energy lost by the water. Therefore:
Q_al = -Q_water
Substituting the values:
m_al * c_al * ΔT_al = - (m_water * c_water * ΔT_water)
Now, you can substitute the given values:
m_al = 32.5g
c_al = 0.903 cal/(g·°C)
ΔT_al = final temperature - initial temperature of aluminum
m_water = 105.3g
c_water = 4.18 cal/(g·°C)
ΔT_water = final temperature - initial temperature of water
Let's denote the final temperature of both substances as T_f.
32.5g * 0.903 cal/(g·°C) * (T_f - 45.8°C) = - (105.3g * 4.18 cal/(g·°C) * (T_f - 15.4°C))
Now, you need to solve this equation for T_f.
First, distribute the negative sign on the right side of the equation:
32.5g * 0.903 cal/(g·°C) * (T_f - 45.8°C) = -105.3g * 4.18 cal/(g·°C) * T_f + 105.3g * 4.18 cal/(g·°C) * 15.4°C
Now, simplify the equation:
32.5g * 0.903 cal/(g·°C) * T_f - 32.5g * 0.903 cal/(g·°C) * 45.8°C = -105.3g * 4.18 cal/(g·°C) * T_f + 105.3g * 4.18 cal/(g·°C) * 15.4°C
32.5g * 0.903 cal/(g·°C) * T_f + 105.3g * 4.18 cal/(g·°C) * T_f = 32.5g * 0.903 cal/(g·°C) * 45.8°C + 105.3g * 4.18 cal/(g·°C) * 15.4°C
Now, solve for T_f by isolating it on one side of the equation:
(32.5g * 0.903 cal/(g·°C) + 105.3g * 4.18 cal/(g·°C)) * T_f = 32.5g * 0.903 cal/(g·°C) * 45.8°C + 105.3g * 4.18 cal/(g·°C) * 15.4°C
Divide both sides of the equation by the sum of the specific heat capacities:
T_f = (32.5g * 0.903 cal/(g·°C) * 45.8°C + 105.3g * 4.18 cal/(g·°C) * 15.4°C) / (32.5g * 0.903 cal/(g·°C) + 105.3g * 4.18 cal/(g·°C))
Now, plug in the given values and calculate T_f using a calculator:
T_f = (32.5g * 0.903 cal/(g·°C) * 45.8°C + 105.3g * 4.18 cal/(g·°C) * 15.4°C) / (32.5g * 0.903 cal/(g·°C) + 105.3g * 4.18 cal/(g·°C))
T_f ≈ 33.4°C
Therefore, the final temperature of both substances at thermal equilibrium is approximately 33.4°C.