What is the final temperature of a system?

Question

What is the final temperature of a system if 10.00 g of gold at 90.0°C is placed in 10.00 g of water at 28.00°C? The molar heat capacity of gold is 25.41 J/(mol · °C) and the heat capacity of water is 4.18 J/(g · °C).

Answer 1

Before beginning, we should convert the specific heats so that they have the same units. I will convert the specific heat for gold from 25.41 J/molº to J/gº to be consistent with the units used for water.

25.41 J / molº x 197.0 g Au / mol = 0.129 J/gº = specific heat for gold

Now, we can set up our equations for the heat LOST by hot gold must equal heat GAINED by cooler water

heat lost by gold = q = mC∆T

q = (10.00 g)(0.129 J/gº)(90.0º - T_f) where T_f = the final temperature

q = 116.1 - 1.29T_f

heat gained by water = q = mC∆T

q = (10.00 g)(4.18 J/gº))(T_f - 28.00º)

q = 41.8T_f - 1170

Setting them equal to each other and solving for T_f we have...

116.1 - 1.29T_f = 41.8T_f - 1170

43.09T_f = 1286

T_f = 29.8ºC