A 3 Kg lead ball is dropped into a 3 L insulated pail of water initially at 20.0 degrees C. If the final temperature of the water-lead combination is 25 degress C, what was the initial temperature of the lead ball?
the sum of the heats gained is zero.
heat gained by water+heat gained by Pb =0
3000*Cw(Tf-20)+3*Cpb*(Tf-Ti)=0
solve for Ti lead.
69 degrees
To solve this problem, we can use the principle of conservation of energy. We know that the heat lost by the lead ball is equal to the heat gained by the water and the lead ball.
The heat lost by the lead ball can be calculated using the formula:
Q = mcΔT,
where Q is the heat lost, m is the mass of the lead ball, c is the specific heat capacity of lead, and ΔT is the change in temperature of the lead ball.
Similarly, the heat gained by the water and the lead ball can be calculated using the formula:
Q = mcΔT,
where Q is the heat gained, m is the total mass of the water and the lead ball, c is the specific heat capacity of water, and ΔT is the change in temperature of the water and the lead ball.
Since the heat lost is equal to the heat gained, we can set up the following equation:
mcΔT (lead) = mcΔT (water + lead).
Given that the mass of the lead ball (m) is 3 kg, the specific heat capacity of lead (c) is 0.13 J/g°C, the specific heat capacity of water (c) is 4.18 J/g°C, the initial temperature of the water (T) is 20.0°C, and the final temperature of the water-lead combination (T) is 25°C, we can solve for the initial temperature of the lead ball (T).
Let's plug in the values:
(3 kg)(0.13 J/g°C)(T - 25°C) = (3 kg + 3 kg)(4.18 J/g°C)(25°C - 20.0°C).
Simplifying the equation:
0.39(T - 25°C) = 24.9(5.0°C).
0.39T - 9.75 = 124.5.
0.39T = 124.5 + 9.75.
0.39T = 134.25.
T ≈ 344.2°C.
Therefore, the initial temperature of the lead ball was approximately 344.2°C.
To find the initial temperature of the lead ball, we can use the principle of heat transfer. The heat gained or lost by an object can be calculated using the equation:
Q = mcΔT
Where:
Q is the heat gained or lost
m is the mass of the object
c is the specific heat capacity of the material
ΔT is the change in temperature
In this case, the lead ball transfers heat to the water, so the heat gained by the water is equal to the heat lost by the lead ball.
Let's consider the heat gained by the water, using the equation:
Q_water = m_water * c_water * ΔT_water
Where:
Q_water is the heat gained by the water
m_water is the mass of the water
c_water is the specific heat capacity of water
ΔT_water is the change in temperature of the water
Since the water is initially at 20.0 degrees C and the final temperature is 25 degrees C, ΔT_water = 25 - 20.0 = 5.0 degrees C.
Now, let's consider the heat lost by the lead ball, using the equation:
Q_lead = m_lead * c_lead * ΔT_lead
Where:
Q_lead is the heat lost by the lead ball
m_lead is the mass of the lead ball (3 kg)
c_lead is the specific heat capacity of lead
ΔT_lead is the change in temperature of the lead ball
We want to find the initial temperature of the lead ball, so we need to rearrange the equation to solve for ΔT_lead:
ΔT_lead = Q_lead / (m_lead * c_lead)
Since the heat gained by the water is equal to the heat lost by the lead ball:
Q_water = Q_lead
Therefore, we can equate the two equations:
m_water * c_water * ΔT_water = m_lead * c_lead * ΔT_lead
Now we can solve for ΔT_lead:
ΔT_lead = m_water * c_water * ΔT_water / (m_lead * c_lead)
Substituting the given values:
m_water = 3 L = 3 kg (since the density of water is approximately 1 g/mL or 1 kg/L)
c_water = 4.18 J/g°C (specific heat capacity of water)
c_lead = 0.13 J/g°C (specific heat capacity of lead)
ΔT_lead = (3 kg) * (4.18 J/g°C) * (5.0°C) / (3 kg) * (0.13 J/g°C)
ΔT_lead ≈ 64.62°C
Finally, we can find the initial temperature of the lead ball by subtracting ΔT_lead from the final temperature:
Initial temperature of the lead ball = Final temperature - ΔT_lead
Initial temperature of the lead ball = 25°C - 64.62°C
Initial temperature of the lead ball ≈ -39.62°C
Therefore, the initial temperature of the lead ball was approximately -39.62°C.