A 4.0 kg gold bar at 97C is dropped into 0.27 kg of water at 22C.
What is the final temperature? Assume the
specific heat of gold is 129 J/kg C.
Answer in units of C.
heat lost by Au bar + heat gained by H2O = 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, we can use the principle of conservation of energy. The heat lost by the gold bar will be equal to the heat gained by the water.
The heat lost by the gold bar can be calculated using the formula:
Q_gold = m_gold * c_gold * ΔT_gold
Where:
Q_gold is the heat lost by the gold bar (in Joules)
m_gold is the mass of the gold bar (4.0 kg)
c_gold is the specific heat of gold (129 J/kg C)
ΔT_gold is the change in temperature of the gold bar
Similarly, the heat gained by the water can be calculated using the formula:
Q_water = m_water * c_water * ΔT_water
Where:
Q_water is the heat gained by the water (in Joules)
m_water is the mass of the water (0.27 kg)
c_water is the specific heat of water (which is 4186 J/kg C)
ΔT_water is the change in temperature of the water
Since the gold bar is dropped into the water, it will heat up the water, resulting in a positive change in temperature. So we'll take ΔT_water as the final temperature of the system.
Since the heat lost by the gold bar equals the heat gained by the water, we can set up the equation:
Q_gold = Q_water
m_gold * c_gold * ΔT_gold = m_water * c_water * ΔT_water
Substituting the given values, we have:
4.0 kg * 129 J/kg C * (97C - ΔT_gold) = 0.27 kg * 4186 J/kg C * (ΔT_water - 22C)
We can solve this equation to find the value of ΔT_water, which will give us the final temperature.
Let's calculate it step by step: