Question
Asked Nov 8, 2019
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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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Expert Answer

Step 1

Given,

Mass of lead weight, mlead = 2.66 g

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

Specific heat of lead, clead = 0.128 J/g⋅°C

Step 2

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

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ATwater Final temperature - Initial temperature of water ATwater = Tr- 52.6 °C ATiead Final temperature Initial temperature of steel ATlead Tf 10.6 °C

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Step 3

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

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Alost gain - (mwater X Cwater ATwater)= (mjead X Clead X ATjead) -[7.77 g x 4.184 J/g.C x (Tr-52.6 °C) 2.66 g x 0.128 J/g-C x (Tf 10.6 °C) 7.77 g x 4.184 J/g°C x (52.6 °C - T¢) 2.66 g x 0.128 J/gC x (Tr 10.6 °C) J 7.77 g x 4.184 в ос x (52.6°C - Т) - (Tr- 10.6 °C) 2.66 g x 0.128 g oC 95.482 x (52.6°C - Tt)= (Tf- 10.6 °C) 5022.3532 °C -95.482 Tr=Tç- 10.6 °C 95.482 Tr+Tf= 5022.3532 °C+ 10.6 °C 5032.9532 c 96.482 Tr 52.2 °C

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