Question

A 2.76 g lead weight, initially at 10.7 ∘C, is submerged in 8.17 g of water...

A 2.76 g lead weight, initially at 10.7 ∘C, is submerged in 8.17 g of water at 52.6 ∘C in an insulated container.

What is the final temperature of both the weight and the water at thermal equilibrium?

Express the temperature in Celsius to three significant figures.

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Answer #1

heat lose of water            =   heat gain of lead

mcT                             =   mcT

8.17*4.184*(52.6-t)          = 2.76*0.128*(t-10.7)

   t   = 52.20C

The final temperature is 52.20C

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