Answer
To solve this problem, we can use the principle of conservation of energy, which states that the total energy of a closed system remains constant. In this case, the closed system consists of the skillet and the water.
The energy transferred from the skillet to the water 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 skillet to the water:
Q1 = m1 * cp1 * ΔT1
Q1 = 2.33 kg * 0.91 kJ/kg°C * (286°C - T)
Where cp1 is the specific heat capacity of aluminum and T is the final temperature of the skillet and water.
The energy gained by the water is:
Q2 = m2 * cp2 * ΔT2
Q2 = m2 * 4.18 kJ/kg°C * (T - 25°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 skillet:
Q1 = Q2
2.33 kg * 0.91 kJ/kg°C * (286°C - T) = m2 * 4.18 kJ/kg°C * (T - 25°C)
Simplifying the equation, we get:
2.33 * 0.91 * (286 - T) = m2 * 4.18 * (T - 25)
Solving for m2, we get:
m2 = 0.54 kg
Therefore, the minimum amount of water required to keep the equilibrium temperature below 40°C is 0.54 kg, or 540 grams.
Work Step by Step
To solve this problem, we can use the principle of conservation of energy, which states that the total energy of a closed system remains constant. In this case, the closed system consists of the skillet and the water.
The energy transferred from the skillet to the water 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 skillet to the water:
Q1 = m1 * cp1 * ΔT1
Q1 = 2.33 kg * 0.91 kJ/kg°C * (286°C - T)
Where cp1 is the specific heat capacity of aluminum and T is the final temperature of the skillet and water.
The energy gained by the water is:
Q2 = m2 * cp2 * ΔT2
Q2 = m2 * 4.18 kJ/kg°C * (T - 25°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 skillet:
Q1 = Q2
2.33 kg * 0.91 kJ/kg°C * (286°C - T) = m2 * 4.18 kJ/kg°C * (T - 25°C)
Simplifying the equation, we get:
2.33 * 0.91 * (286 - T) = m2 * 4.18 * (T - 25)
Solving for m2, we get:
m2 = 0.54 kg
Therefore, the minimum amount of water required to keep the equilibrium temperature below 40°C is 0.54 kg, or 540 grams.
