Question

A 2.04 g lead weight, initially at 10.8 oC, is submerged in 7.62 g of water at 52.3 oC in an insulated container. clear = 0.128 J/g oF; water = 4.18 J/goC. What is the final temperature of both the weight and the water at thermal equilibrium

Answer 1

Answer: The final temperature of both the weight and the water at thermal equilibrium is $50.26^{o}C$.

Explanation:

The given data is as follows.

mass = 7.62 g,           $T_{2} = 10.8^{o}C$

Let us assume that T be the final temperature. Therefore, heat lost by water is calculated as follows.

       q = $mC \times \Delta T$    

          = $7.62 g \times 4.184 J/^{o}C \times (52.3 - T)$

Now, heat gained by lead will be calculated as follows.

       q = $mC \times \text{Temperature change of lead}$  

           = $2.04 \times 0.128 \times (T - 11.0)$

According to the given situation,

     Heat lost = Heat gained

$7.62 g \times 4.184 J/^{o}C \times (52.3 - T)$ = $2.04 \times 0.128 \times (T - 11.0)$

        T = $50.26^{o}C$

Thus, we can conclude that the final temperature of both the weight and the water at thermal equilibrium is $50.26^{o}C$.