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14 August, 06:25

A 45-g aluminum spoon (specific heat 0.88 j/g °c) at 24 °c is placed in 180 ml (180 g) of coffee at 85 °c andthe temperature of the two become equal. (a) what is the final temperature when the two become equal? assume that coffee has the same specific heat aswater.

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  1. 14 August, 07:12
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    When Heat lost by coffee - q = m (coffee) * C (coffee) * ΔT

    and Heat gained by spoon q = m (spoon) * C (spoon) * ΔT

    when q = - q

    ∴ m (coff) * C (coff) * ΔT (coff) = m (sp) * C (sp) * ΔT (sp)

    when the final tempreature of both will be the same so,

    m (coff) * C (coff) * (Tf - Ti (coff)) = m (sp) * C (sp) * (Tf - Ti (sp)

    by substitution:

    note: as the coffe will reduce heat so, ΔT = (85-Tf)

    and the spoon will gain heat so ΔT = (Tf-24)

    180 g * 4.186 * (85-Tf) = 45 * 0.88 * (Tf - 24)

    ∴Tf = 81 °C
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