Answer:
The temperature of the water and lead weight is 31.0°C.
Step-by-step explanation:
To solve this problem, we can use the formula:
q = mcΔT
where q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature.
First, let's calculate the heat transferred from the water to the lead weight:
q1 = mcΔT = (7.45 g)(4.184 J/g°C)(52.2°C - T) where T is the final temperature of the water and lead weight
q2 = mcΔT = (2.26 g)(0.128 J/g°C)(T - 11.1°C)
Since the container is insulated, the heat transferred from the water to the lead weight is equal to the heat transferred from the lead weight to the water:
q1 = q2
(7.45 g)(4.184 J/g°C)(52.2°C - T) = (2.26 g)(0.128 J/g°C)(T - 11.1°C)
Simplifying and solving for T:
T = 31.0°C
Therefore, the final temperature of the water and lead weight is 31.0°C.