Question

A 2.04 g lead weight, initially at 10.8 oC, is submerged in 7.74 g of water at 52.2 oC in an insulated container. clead = 0.128 J/goC; cwater = 4.18 J/goC. What is the final temperature of both the weight and the water at thermal equilibrium?

Answer 1

Answer:

Explanation:

Hello, in this case, the lead is catching heat and the water losing it, that's why the heat relation ship is (D is for Δ):

$DH_{lead}=-DH_{water}$

Now, by stating the heat capacity definition:

$m_{Pb}C_{Pb}*(T_{eq}-T_{lead}=-m_{H_2O}C_{H_2O}*(T_{eq}-T_{H_2O})$

Solving for the equilibrium temperature:

$T_{eq}=\frac{m_{Pb}C_{Pb}T_{Pb}+m_{H_2O}C_{H_2O}T_{H_2O}}{m_{Pb}C_{Pb}+m_{H_2O}C_{H_2O}} \\\\T_{eq}=\frac{2.04g*0.128J/(g^oC)*10.8^oC+7.74g*4.18J/(g^oC)*52.2^oC}{2.04g*0.128J/(g^oC)+7.74g*4.18J/(g^oC)} \\\\T_{eq}=51.87^oC$

Which is very close to the water's temperature since the lead's both mass and head capacity are lower than those for water.

Best regards.