What would be the final temperature of the
system if 28.4 g of lead at 99.5 ◦C is dropped
into 14.0 g of water at 8.61 ◦C in an insulated
container? The specific heat of lead is 0.128
J/g◦C.
Jane -- I can't help you -- except to say this looks like a homework dump. Many of our tutors stay far away from a bunch of questions posted by the same person.
I agree with Ms. Sue. If you will start at the first post, tell us what you think about each problem, tell us EXACTLY what you don't understand about the problem and how much you know how to do on your own, I'll be more than happy to help through each and every one.
This will help you through this one. The heat given up by lead to the water must equal to the heat absorbed by the water wo there is zero change. It looks this way.
[mass Pb x specific heat Pb x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
You have mass Pb, specific heat Pb, you're look for Tfinal,you have Tinitial Pb and Tinitial H2O. I assume you know specific heat H2O.
Post your work if you get stuck.
To find the final temperature of the system, we can use the principle of heat exchange, which states that the heat lost is equal to the heat gained.
First, we need to calculate the heat lost by the lead. The heat lost by an object can be found using the formula:
Q = m × c × ΔT
where Q is the heat lost, m is the mass of the object, c is the specific heat capacity, and ΔT is the change in temperature.
For the lead, the mass is 28.4 g, the specific heat capacity is 0.128 J/g◦C, and the change in temperature is the final temperature minus the initial temperature (Tf - Ti). The initial temperature is 99.5 ◦C, and since the system is insulated, we assume that the final temperature is the same for both the lead and water.
Now, let's calculate the heat lost by the lead:
Q_lead = m_lead × c_lead × (T_lead_f - T_lead_i)
= 28.4 g × 0.128 J/g◦C × (Tf - 99.5 ◦C)
Next, we need to calculate the heat gained by the water.
Q_water = m_water × c_water × (T_water_f - T_water_i)
= 14.0 g × 4.18 J/g◦C × (Tf - 8.61 ◦C)
According to the principle of heat exchange, Q_lead = Q_water.
Substituting the values and equating the two equations, we get:
28.4 g × 0.128 J/g◦C × (Tf - 99.5 ◦C) = 14.0 g × 4.18 J/g◦C × (Tf - 8.61 ◦C)
Now, we can solve for Tf.
Multiply out the equation and isolate Tf:
3.6352 × (Tf - 99.5 ◦C) = 58.52 × (Tf - 8.61 ◦C)
3.6352 Tf - 360.896 ◦C = 58.52 Tf - 491.005
Combine like terms:
58.52 Tf - 3.6352 Tf = -360.896 ◦C + 491.005
Simplify:
54.8848 Tf = 130.109
Finally, isolate Tf:
Tf = 130.109 / 54.8848
Thus, the final temperature of the system would be Tf = 2.367 ◦C, rounded to three decimal places.