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A gem dealer wants to find the specific heat of 0,025 kg of precious stones?

Sample is heated to 95.0 ° C, then placed at 0.13 kg copper vessel containing a 0.089 kg of water in equilibrium at 25.0 ° C. The heat loss to the external environment is negligible. When equilibrium is established, temperature is 27.0 ° C. What is the specific heat capacity of the sample?

Cu = Weight 65.5g/mol. . . . (From Wikipedia) The heat capacity Copper = 24.440 J / (mol K). . . (From Wikipedia) =. 3725 J / (g • K) heat capacity of water = 4.184 J / (g • K). . . (From Wikipedia) Variation Water Energy Thermal Energy Kettle + + gems Change = 0 (27.0 º C and 25.0 ° C) * (Cu 130 g) * (0.3725 J / (G • K) + (27.0 º C and 25.0 ° C) * (89 g H20) * (4.184 J / (K • g)) + (27.0 º C and 95.0 ° C) * (25 g Heat) * Ability Gemstone = 0 therefore the heat capacity of the gemstones = – [(27.0 º C and 25.0 ° C) * (Cu 130 g) * (0.3725 J / (g • K) + (27.0 C- 25.0 ° C) * (89 g H20) * (4.184 J / (K • g))] / (27.0 º C and 95.0 ° C) * (25 g), =. 495 J / (g ° K)

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