Question

A 450g cube of gold at 225 degrees Celsius is cooled by dunking it in 850g of water at 20 degrees C. what is the final temperature of the mixture?
specific heat of gold = 0.031 cal/gC
specific heat of water = 1

Answer 1

q lost = q gain
So, by substitution, we then have:
(450) (225 - x)(0.031) = (850) (x - 20) (1.0)
Solve for x
13.95 (225 - x) = 850 (x - 20)
3138.75 - 13.95x = 850x - 17000
then 863.95x = 20138.75
The answer is 23.31 °C

Answer 2

U know heat loss by gold is just equal to the heat gained by water

Answer 3

We assume this water and gold r isolated frm environment

Answer 4

So no heat transfer will occur between the system ( water n gold ) and environment

Answer 5

Here used
Q=ms*del theta

Answer 6

Q= heat loss or heat gain

Answer 7

m=mass of object

Answer 8

S=specific heat of the object

Answer 9

Del theta= temperature difference