What would be the final temperature of the

system if 20 g of lead at 100◦C are dropped
into 20 g of water at 30 ◦C in an insulated
container?

heat lost by lead + heat gained by water = 0

[mass Pb x specific heat Pb x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute all the number and solve for Tfinal. That's the only unknown in that long string.

To determine the final temperature of the system, we can use the principle of conservation of energy, specifically the principle of heat transfer.

First, we need to calculate the heat transfer between the lead and water. The heat transfer equation is given by:

Q = m * c * ΔT

Where:
Q is the heat transfer
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature

For lead:
m = 20 g (given)
c = 0.13 J/g °C (specific heat capacity of lead)
ΔT = Tfinal - Tinitial

For water:
m = 20 g (given)
c = 4.18 J/g °C (specific heat capacity of water)
ΔT = Tfinal - Tinitial

Since the container is insulated, we can assume that no heat is lost to the surroundings. Therefore, the heat gained by the water is equal to the heat lost by the lead.

Qwater = -Qlead

Now we can rewrite the equation using the equations above:

mwater * cwater * ΔTwater = -mlead * clead * ΔTlead

Substituting the given values:

20 g * 4.18 J/g °C * (Tfinal - 30°C) = -20 g * 0.13 J/g °C * (Tfinal - 100°C)

Simplifying the equation gives:

4.18 Tfinal - 4.18 * 30 = -0.13 Tfinal + 0.13 * 100

4.18 Tfinal - 125.4 = -0.13 Tfinal + 13

Adding 0.13 Tfinal and subtracting 13 from both sides gives:

4.18 Tfinal + 0.13 Tfinal = 125.4 + 13

4.31 Tfinal = 138.4

Dividing both sides by 4.31 gives:

Tfinal ≈ 32.08°C

Therefore, the final temperature of the system would be approximately 32.08°C.