A steel casing weighing 2 kg has an initial temperature of 500°C;
40 kg of water initially at 25°C is contained in a perfectly
insulated steel tank weighing 5 kg. The casing is immersed in
water and the system is allowed to come to equilibrium. What is
the final temperature? Ignore the effects of expansion and
contraction, and assume constant specific heats for water and steel
to be 4.18 kJ/kg-K and 0.50 kJ/kg-K.
To solve this problem, we can use the concept of heat transfer. The heat transferred from the steel casing to the water will be equal to the heat gained by the water.
Let's first calculate the initial heat content of the steel casing:
Q_initial = m_initial * c_steel * ΔT_initial
Q_initial = 2 kg * 0.50 kJ/kg-K * (500°C - 25°C)
Q_initial = 2 kg * 0.50 kJ/kg-K * 475°C
Q_initial = 475 kJ
Now, let's calculate the initial heat content of the water:
Q_initial = m_initial * c_water * ΔT_initial
Q_initial = 40 kg * 4.18 kJ/kg-K * (T_final - 25°C)
Q_initial = 166.8 kJ/kg-K * (T_final - 25°C)
Since the system is perfectly insulated, the total heat content of the system remains constant:
Q_initial + Q_initial = Q_final
475 kJ + 166.8 kJ/kg-K * (T_final - 25°C) = 0
Now, let's solve for T_final:
475 kJ + 166.8 kJ/kg-K * T_final - 4167 kJ = 0
166.8 kJ/kg-K * T_final = 4167 kJ - 475 kJ
166.8 kJ/kg-K * T_final = 3692 kJ
T_final = 3692 kJ / (166.8 kJ/kg-K)
T_final = 22.11°C
Therefore, the final temperature of the system is approximately 22.11°C.