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13 December, 08:53

A 2.77 g lead weight, initially at 11.1 oC, is submerged in 7.94 g of water at 52.8 oC in an insulated container. clead = 0.128 J/goC; cwater = 4.18 J/goC. What is the final temperature of both the weight and the water at thermal equilibrium?

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  1. 13 December, 09:19
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    52.36 ºC

    Explanation:

    By the law of the conservation of energy, the heat lost by the water must equal the heat gained by lead.

    q H₂O = q Pb

    Now q is given by m x c x ΔT

    where m is the mass, c the specific heat, and ΔT the change in temperature.

    So plugging our values in the eaquation above:

    q H₂O = q Pb

    7.94 g x 4.18 J/gºC x (52.8 ºC - T₂) = 2.77 g x 0.128 J/gºC x (T₂ - 11.1 ºC)

    1752.390 - 33.189 T₂ = 0.355 T₂ - 3.936

    33.544 T₂ = 1756.326

    T₂ = 52.36 ºC

    Notice that the change in temperature for water is very small, a reflection of the very large specific heat of water.
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