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5. A 23.9 g piece of metal heated to 97.8 °C is placed in 52.4 g water at 21.9 °C. After the metal is added, the temperature
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Answer #1

- Heat lost by the metal = Heat gained by the water

-Qlost = Qgained

Negative sign indicates the loss of heat

-Qm = Qw

Mm x Cm x dt = Mw x Cw x dt

Mm x Cm x (tf-ti) = Mw x Cw x (tf-ti)

- 23.9 g x Cm x (29.9 - 97.8) = 52.4 g x 1.00 Cal/g C x (29.9-21.9)

23.9 x 67.9 x Cm = 52.4 x 8

Cm = 419.2 Cal/g C/1622.81 = 0.258 Cal/g C

Specific heat of metal = 0.258 Cal/g C

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