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

2 answers

The sum of the heats gained is zero (something loses heat).

Heatgainedwater+heatgainedgold=0
.22*Cw*(Tf-25)+3*Cg*(Tf-99)=0
solve for Tf
A*t =v-u at=v-u add u to both side at*u=v-at