physics
posted by Jim on .
What is the final tempurature when a 3.0 kg gold bar at 99C is dropped into 0.22 kg of water at 25C Specific heat of gold is 129 j/kg x C

The sum of the heats gained is zero (something loses heat).
Heatgainedwater+heatgainedgold=0
.22*Cw*(Tf25)+3*Cg*(Tf99)=0
solve for Tf 
A*t =vu at=vu add u to both side at*u=vat