Question

9. You drop a 15.0g sample of an unknown metal at 95.0oC into a 45.5g sample of H2O(l) at 19.2oC. When the mixture reaches thermal equilibrium the temperature of the metal and the H2O(l) is 24.2oC. What is the specific heat capacity of the metal? (The Cs of H2O(l) is 4.18J/goC.)

Answer 1

The energy lost by unknow = The energy gained by water. Mass . Metal of x Cs, metal X Change in - Metal Temp. Metal Temp. Mass of X water w Cs, water & change in water temp. temp. 15.09 x Cs, mutol (95-24.2°C) = 45.59 x 4:19 Ix (24.2-1:20) 9.0c 1062 Cs, metal = 950.95 Cs, metal = 0.895 J Specific heat is 0.895 J capacity of unknown metal