A small “coffee cup” calorimeter contains 110. g of H2O at 22.0°C. A 100. g sample of lead is heated to 90.0°C and then placed in the water. The contents of the calorimeter come to a temperature of 23.9°C What is the specific heat of lead?

heat lost by metal + heat gained by water = 0

[mass lead x specific heat lead x (Tfinal-Tinitial)] + [mass water x specific heat water x (Tfinal-Tinitial)]=0
Substitute the numbers and solve for the one unknown in the equation.

To find the specific heat of lead, we need to use the principle of energy conservation and the equation Q₁ + Q₂ = 0, where Q₁ is the heat gained by the water and Q₂ is the heat lost by the lead.

The equation for heat transfer is Q = mcΔT, where Q is the heat gained or lost, m is the mass, c is the specific heat, and ΔT is the change in temperature.

First, let's calculate the heat gained by the water. The mass of water is 110 g, the specific heat of water is approximately 4.18 J/g°C, and the change in temperature (ΔT) is the final temperature (23.9°C) minus the initial temperature (22.0°C):

Q₁ = (110 g) * (4.18 J/g°C) * (23.9°C - 22.0°C)

Next, let's calculate the heat lost by the lead. The mass of lead is 100 g, and the change in temperature (ΔT) is the final temperature (23.9°C) of the calorimeter minus the temperature of the lead (90.0°C):

Q₂ = -(100 g) * (c lead) * (23.9°C - 90.0°C)

Since the heat gained by the water and the heat lost by the lead are equal (Q₁ + Q₂ = 0), we can set up the equation:

(110 g) * (4.18 J/g°C) * (23.9°C - 22.0°C) = -(100 g) * (c lead) * (23.9°C - 90.0°C)

Now, let's solve this equation for the specific heat of lead (c lead).