A hot stell ball of mass 1 kg is dropped into a 4 L bucket of water at room temperature (70 degrees F). If the system reaches an equilibrium temperature of 320 K, what was the initial temperature of the steel ball before it was dropped into the water? Cp of steel = 490 J/kg degrees C, Cp of water = 4186 J/kg degrees C.
the sum of heats gained is zero.
HeatgainedSteelBall+ HeatgainedWater=0
1kg*490*(Tf-Ti)+4000*4186*(Tf-70)=0
solve for Ti
thank you!
To find the initial temperature of the steel ball before it was dropped into the water, we can use the principle of conservation of energy.
The heat gained by the water is equal to the heat lost by the steel ball. The formula to calculate the heat gained or lost is:
Q = m * Cp * ΔT
Where:
Q is the amount of heat gained or lost
m is the mass of the substance
Cp is the specific heat capacity of the substance
ΔT is the change in temperature
Let's assume the initial temperature of the steel ball is T1.
The heat gained by the water is given by:
Q_water = m_water * Cp_water * (T_equilibrium - T_room)
Where:
m_water = 4 kg (since the volume is given as 4 L, and assuming the density of water is 1 kg/L)
Cp_water = 4186 J/kg degrees C (specific heat capacity of water)
T_equilibrium = 320 K (equilibrium temperature)
T_room = 70 degrees F = 21.1 degrees C (room temperature)
The heat lost by the steel ball is given by:
Q_steel = m_steel * Cp_steel * (T1 - T_equilibrium)
Where:
m_steel = 1 kg (mass of the steel ball)
Cp_steel = 490 J/kg degrees C (specific heat capacity of steel)
Since the total heat gained by the water is equal to the total heat lost by the steel ball, we equate the two equations:
Q_water = Q_steel
m_water * Cp_water * (T_equilibrium - T_room) = m_steel * Cp_steel * (T1 - T_equilibrium)
Plugging in the given values:
4 * 4186 * (320 - 21.1) = 1 * 490 * (T1 - 320)
Simplifying:
4 * 4186 * 298.9 = 1 * 490 * (T1 - 320)
Solving for T1:
(4 * 4186 * 298.9) / (1 * 490) + 320 = T1
T1 = 730.12 degrees C
Therefore, the initial temperature of the steel ball before it was dropped into the water was approximately 730.12 degrees Celsius.