# A 2.66 g lead weight, initially at 10.6 ∘C, is submerged in 7.77 gof water at 52.6 ∘C in an insulated container.What is the final temperature of both the weight and the water at thermal equilibrium?

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A 2.66 g lead weight, initially at 10.6 ∘C, is submerged in 7.77 gof water at 52.6 ∘C in an insulated container.

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

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Step 1

Given,

Mass of water, mwater = 7.77 g

Initial temperature of water, Twater = 52.6 °C

Initial temperature of steel, Tlead = 10.6 °C

Final temperature, Tf = ?

To solve this we need,

Specific heat of water, cwater = 4.184 J/g⋅°C

Step 2

The change in temperature for water and lead can be calculated as:

Step 3

The heat lost by water (– qlost) when lead weight is submerged in the water will be equal...

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