A 250 gram sample of metal shot is heated to a temperature of 98oC. It is placed in 100.0 grams of water in a brass calorimeter cup with a brass stirrer. The total mass of the cup and the stirrer is 50.0 grams. The initial temperature of the water, stirrer, and calorimeter cup is 20.0oC. The final equilibrium temperature of the system is 30.0oC. What is the specific heat of the metal sample?
Here also q lost = q gained
Heat lost by metal = -q lost = -ms (tf-ti) = 250 gram .x.(98-30)
q gained = ms(tf-ti)water + ms(tf-ti)brass
=100.0 grams*4.18*(30-20) + 50*0.385(30-20)
-250 gram .x.(98-30) =100.0 grams*4.18*(30-20) + 50*0.385(30-20)
solve it for x