A sample of gold is at a initial temperature of 350 k. Once dropped into a coffe cup calorimeter of heat capacity 400 k^-1 it contains 50 g of water at 300 k. Specific heat capacity is 0.129 for gold and 4.18 for water. what is the final temp of all gold, coffee cup, and water.
heat lost by Au + heat gained by water = 0
[mass Au x specific heat Au x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Solve for Tfinal.
To find the final temperature of the gold, coffee cup, and water, we can use the concept of heat transfer.
The heat lost by the gold will be equal to the heat gained by the water and the coffee cup. We can use the formula:
Q = mcΔT
where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Let's calculate the heat transferred by the gold:
Q_gold = mgΔT_gold
Q_gold = (50 g) (0.129 k^-1) (350 K - T_final)
Now, let's calculate the heat gained by the water and the coffee cup:
Q_water = mcΔT_water
Q_water = (50 g) (4.18 k^-1) (T_final - 300 K)
Q_cup = mcΔT_cup
Q_cup = (400 k^-1) (T_final - 300 K)
Since heat is conserved, we can set up the following equation:
Q_gold = Q_water + Q_cup
Substituting the values, we have:
(50 g) (0.129 k^-1) (350 K - T_final) = (50 g) (4.18 k^-1) (T_final - 300 K) + (400 k^-1) (T_final - 300 K)
Simplifying the equation, we get:
(6.45 K - 0.129 K T_final) = (209 k^-1 T_final - 1254 K) + (1.6 k^-1 T_final - 480 K)
Combine like terms:
(0.129 K T_final + 209 k^-1 T_final + 1.6 k^-1 T_final) = (-6.45 K + 1254 K + 480 K)
(1.929 K T_final) = (1728.55 K)
Divide by 1.929 K:
T_final = 896.54 K
Therefore, the final temperature of the gold, coffee cup, and water is approximately 896.54 K.