Asked by Holly
A 599 g piece of silver at 150 degrees C is submerged in 1000 g of water at 5 degrees C to be cooled. Determine the final temperature of he silver and water given
Heat capacity of water= 4.18 x 10^3 j/kg degrees C
Heat capacity of silver= 2.49 x 10^2 j/kg degrees C
Heat capacity of water= 4.18 x 10^3 j/kg degrees C
Heat capacity of silver= 2.49 x 10^2 j/kg degrees C
Answers
Answered by
bobpursley
The sum of the heats gained is zero.
heatgainedbywater+heatgainedbysilver=0
1000*cw(Tf-5)+599(cs)(Tf-150)=0
solve for Tf
heatgainedbywater+heatgainedbysilver=0
1000*cw(Tf-5)+599(cs)(Tf-150)=0
solve for Tf