Answer
The energy transferred from the fuel rods to the water in the spent fuel pool is given by the formula:
Q = mcpΔT
Where Q is the amount of heat transferred, m is the mass of the object, cp is the specific heat capacity, and ΔT is the change in temperature.
Let's first calculate the amount of heat transferred from the fuel rods to the water:
Q1 = m1 * cp1 * ΔT1
Q1 = 248 * 322 kg * 284 J/kg·K * (658°C - T)
Where cp1 is the specific heat capacity of the fuel rods and T is the final temperature of the fuel rods and water.
The energy gained by the water is:
Q2 = m2 * cp2 * ΔT2
Q2 = 1720 * $10^3 $kg * 4.18 J/kg·K * (T - 15°C)
Where cp2 is the specific heat capacity of water.
Since there is no heat loss, the energy gained by the water is equal to the energy lost by the fuel rods:
Q1 = Q2
248 * 322 kg * 284 J/kg·K * (658°C - T) = 1720 * $10^3 $kg * 4.18 J/kg·K * (T - 15°C)
Simplifying the equation, we get:
248 * 322 * 284 * (658 - T) = 1720 * $10^3$ * 4.18 * (T - 15)
Solving for T, we get:
T = 23.8°C
Therefore, the equilibrium temperature of the fuel rods and water is 23.8°C. However, this temperature will increase further due to the energy generated by radioactive decay in the fuel rods.
Work Step by Step
The energy transferred from the fuel rods to the water in the spent fuel pool is given by the formula:
Q = mcpΔT
Where Q is the amount of heat transferred, m is the mass of the object, cp is the specific heat capacity, and ΔT is the change in temperature.
Let's first calculate the amount of heat transferred from the fuel rods to the water:
Q1 = m1 * cp1 * ΔT1
Q1 = 248 * 322 kg * 284 J/kg·K * (658°C - T)
Where cp1 is the specific heat capacity of the fuel rods and T is the final temperature of the fuel rods and water.
The energy gained by the water is:
Q2 = m2 * cp2 * ΔT2
Q2 = 1720 * $10^3 $kg * 4.18 J/kg·K * (T - 15°C)
Where cp2 is the specific heat capacity of water.
Since there is no heat loss, the energy gained by the water is equal to the energy lost by the fuel rods:
Q1 = Q2
248 * 322 kg * 284 J/kg·K * (658°C - T) = 1720 * $10^3 $kg * 4.18 J/kg·K * (T - 15°C)
Simplifying the equation, we get:
248 * 322 * 284 * (658 - T) = 1720 * $10^3 $* 4.18 * (T - 15)
Solving for T, we get:
T = 23.8°C
Therefore, the equilibrium temperature of the fuel rods and water is 23.8°C. However, this temperature will increase further due to the energy generated by radioactive decay in the fuel rods.
