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.
[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.
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).