If the water had lost 12000 joules and the calorimeter gained 1300 joules as the temperature of both rock and calorimeter increased from 22°C to 73°C and the mass of rock in the calorimeter was 185 grams, what is the specific heat, c, of the rock material?
Q1=Q2+cm•Δt
c=(Q1-Q2)/ m•Δt=
=(12000-1300)/0.185•(73-22)=
=1134 J/kg•℃