A 2.43 g lead weight, initially at 10.6 ∘C, is submerged in 7.78 g of water...

A 2.43 g lead weight, initially at 10.6 ∘C, is submerged in 7.78 g of water at 52.9 ∘C in an insulated container. What is the final temperature of both the weight and the water at thermal equilibrium?

Solutions

Expert Solution

Mass of Pb = 2.43 g

Initial temperature = 10.6^oC

Mass of water = 7.78 g

Temperature of water = 52.9^oC

Heat released by water = m c T
Heat released by water = 0.00778 x 4186 x (52.9 – T_f)

Heat absorbed by lead = m c T
Heat absorbed by lead = 0.00243 x 128 x (T_f – 10.6)

Heat released by water = Heat absorbed by lead

0.00778 x 4186 x (52.9 – T_f) = 0.00243 x 128 x (T_f – 10.6)

(0.00778 x 4186 x 52.9) – (0.00778 x 4186 x T_f) = (0.00243 x 128 x T_f) – (0.00243 x 128 x 11.1)

Add (0.00778 * 4186 * Tf) to both sides

(0.00778 x 4186 x 52.9) = (0.00778 x 4186 x T_f) + (0.00243 x 128 x T_f) – (0.00243 x 128 x 10.6)

Add (0.00243 x 128 x 10.6) to both sides
(0.00243 x 128 x 10.6) + (0.00778 x 4186 x 52.9) = (0.00778 x 4186 x T_f) + (0.00243 x 128 x T_f)

Divide both sides by (0.00778 x 4186) + (0.00243 x 128)

[(0.00243 x 128 x 10.6) + (0.00778 x 4186 x 52.9)] ÷ [(0.00778 x 4186) + (0.00243 x 128)] = Tf

T_f = 52.49^oC

queen_honey_blossom answered 11 months ago

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