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A 34.44-g sample of metal is heated to 98.6 degrees Celsius in a hot water bath until thermal equilibrium is reached What is the specific heat of the metal?
Wiki User
∙ 11y ago
Heat lost by the metal = heat gained by the water.
Heat gained by the water = 50.0 g x 4.184 J/g K x (28.3-22.2) = 1276 J
Heat lost by metal = 1276 J = 34.44 g x Sp Heat x (98.6 - 28.3)
Specific Heat = 1276 J / 2421 g K = 0.527 J/g K
Wiki User
∙ 11y ago
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