Question

A 37.5 g piece of gold at 83.0 �C is added to 100.0 g H2O at 22.0 �C in a well-insulated cup. What is the temperature after the system comes to equilibrium? (The specific heat capacity of Au is 0.129 J�g^�1�K^�1)

Answer 1

Let us find the heat absorbed by water

Q_water = m * C * Delta T

where m = mass of water, 100.0 g

C = specific heat of water, 4.18 J/g C

Let's say final temperature of the system is x.

Therefore delta T for water would be x - 22.0

Substituting the values, we get

Q_water = 100.0 g * 4.18 J/g C * ( x - 22.0 C)

Q_water = 418 ( x - 22.0 ) ....................(1)

Let's find the heat evolved by piece of gold

Q_gold = m * C * Delta T

m for gold is 37.5 g

C is given as 0.129 J/ g K

Delta T for gold is x - 83.0

Q_gold = 37.5 g * 0.129 J/g K * (x - 83)

Q_gold = 4.8375 ( x - 83 ) ......................(2)

Assuming no heat is lost to the surroundings, we have

Q _water = - Q_gold

418 ( x - 22.0 ) = - 4.8375 ( x - 83 )

418 x - 9196 = - 4.8375 x + 401.5125

418 x + 4.8375 x = 401.5125 + 9196

422.8375 x = 9597.5125

x = 9597.5125 / 422.8375

x = 22.7 C

The temperature after the system comes to equilibrium is 22.7 degree C