A precious-stone dealer wishes to find the specific heat capacity of a 0.025 kg gemstone?

Question

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

Answer 1

Weight Cu = 65.5g/mol . . . .(from Wikipedia)
Heat Capacity Copper = 24.440 J/(mol·K) . . . (from Wikipedia)
=.3725 J/(g·K)

Heat Capacity Water = 4.184 J/(g·K) . . . (from Wikipedia)

Change in Heat energy for Water + Kettle + Change in Energy for gemstone = 0

(27.0°C-25.0°C) * (130 g cu) *(.3725 J/(g·K) + (27.0°C-25.0°C) * (89 g h20) * (4.184 J/(g·K)) + (27.0°C-95.0°C) * (25 g) * Heat Capacity gemstone = 0

Therefore

Heat Capacity gemstone = - [(27.0°C-25.0°C) * (130 g cu) *(.3725 J/(g·K) + (27.0°C-25.0°C) * (89 g h20) * (4.184 J/(g·K)) ] / (27.0°C-95.0°C) * (25 g)

=.495 J/(g·K)