A3.5 kg gold bar at 94°c is dropped into 0.20 kg of water at 22°c what is the final temperature? assume the specific heat of gold is 129 j/kg c answer in units of °c

47.17 degree C

Explanation:

mg = 3.5 kg, T1 = 94 degree C, sg = 129 J/kg C

mw = 0.2 kg, T2 = 22 degree C, sw = 4200 J/kg C

Let T be the temperature at equilibrium.

Heat given by the gold = Heat taken by water

mg x sg x (T1 - T) = mw x sw x (T - T2)

3.5 x 129 x (94 - T) = 0.2 x 4200 x (T - 22)

42441 - 451.5 T = 840 T - 18480

60921 = 1291.5 T

T = 47.17 degree C

this would be about 43%