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Bess [88]
4 years ago
7

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
Chemistry
1 answer:
alisha [4.7K]4 years ago
7 0

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.

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