A155.0 g piece of copper at 168 oc is dropped into 250.0 g of water at 20.9 oc. (the specific heat of copper is 0.385 j/goc.) calculate the final temperature of the mixture. (assume no heat loss to the surroundings.)

Note that
The specific heat of water is 4.184 J/(g -C)

Let x = the final equilibrium temperature.

Heat loss by the copper is
(155 g)*(0.385 J/(g -C))*(168 - x C) = 59.675(168 -x) J

Heat gain by the water is
(250 g)*(4.184 J/(g-C))*(x - 20.9 C) = 1046(x - 20.9) J

Because there is no heat loss to the surroundings, therefore
59.675(168 - x) = 1046(x - 20.9)
168 - x = 17.5283(x - 20.9)
18.5283x = 534.3415
x = 28.84 °C

  28.8 °°C  nearest tenth)

10 is the . . .