heat lost by metal + heat gained by water = 0
[mass metal x specific heat metal x (Tfinal-Tinitial)] + [mass water x specific heat water x (Tfinal-Tinitial)] = 0
Substitute and solve for specific heat metal.
A 23.0-g sample of an unknown metal is heated to 91.5°C and is placed in a perfectly insulated container along with 102.g of water at an initial temperature of 21.80°C. After a short time the temperature of both the metal and water become equal at 23.80°C.
The specific Heat Capacity of water is 4.18 J/g/K in this temperature range.
What is the specific heat capacity of the metal?
1 answer
1 answer