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A 0.350 kg block at -27.5°C is added to 0.217 kg of water at 25.0°C. They come to equilibrium at 16.4°C. What is the specific heat of the block?

A 0350 kg block at 275C is added to 0217 kg of water at 250C They come to equilibrium at 164C What is the specific heat of the block class=

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jay40891

Answer:

508 J/kg/C

Explanation:

Energy Lost by water = Energy gained by block

mcT = mcT [bolded is for water, italicised is for block]

(0.217)(4186)(25 - 16.4) = (0.350)(c)(16.4 + 27.5)

15.365c = 7811.9132

c = 508 J/kg/C (3 sf)

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