Equal 0.400-kg masses of lead and tin at 60.0°C are placed in 1.00 kg of water at 20.0°C. If an alloy is half lead and half tin by mass, what specific heat would you anticipate for the alloy?
To solve this problem, we can use the principle of energy conservation.
First, let's calculate the total energy gained by the water. We can use the equation:
Q_water = m_water * c_water * ΔT
where:
- Q_water is the energy gained by the water
- m_water is the mass of the water (1.00 kg)
- c_water is the specific heat of water (4.18 J/g°C)
- ΔT is the change in temperature of the water (final temperature - initial temperature)
ΔT = final temperature - initial temperature
= Tf - Ti
= 60.0°C - 20.0°C
= 40.0°C
Q_water = (1.00 kg) * (4.18 J/g°C) * (40.0°C)
= 167.2 J
Next, let's calculate the total energy lost by the lead and tin. Since both masses are equal and half of the alloy, we can calculate the energy lost by one of the metals and then multiply it by 2.
Q_lead or tin = m_lead or tin * c_lead or tin * ΔT
where:
- Q_lead or tin is the energy lost by lead or tin
- m_lead or tin is the mass of lead or tin (0.400 kg)
- c_lead or tin is the specific heat of lead or tin (to be determined)
- ΔT is the change in temperature of lead or tin (final temperature - initial temperature)
Q_lead or tin = (0.400 kg) * c_lead or tin * ΔT
Since the alloy is half lead and half tin by mass, the total energy lost by lead and tin is:
Q_total = (Q_lead or tin) + (Q_lead or tin)
= 2 * (0.400 kg) * c_lead or tin * ΔT
= 0.800 kg * c_lead or tin * ΔT
Finally, using the principle of energy conservation, the energy gained by the water is equal to the total energy lost by the lead and tin:
Q_water = Q_total
167.2 J = 0.800 kg * c_lead or tin * ΔT
Substituting the value of ΔT:
167.2 J = 0.800 kg * c_lead or tin * 40.0°C
Simplifying the equation:
167.2 J = 32.0 kg°C * c_lead or tin
Dividing both sides by 32.0 kg°C:
c_lead or tin = 167.2 J / 32.0 kg°C
c_lead or tin = 5.225 J/g°C
Therefore, the specific heat anticipated for the alloy is 5.225 J/g°C.
To find the specific heat of the alloy, we can use the principle of conservation of energy. The heat lost by the lead and tin will be gained by the water. We can use the equation:
Q_lead + Q_tin = Q_water
where Q_lead is the heat lost by the lead, Q_tin is the heat lost by the tin, and Q_water is the heat gained by the water.
The formula for the heat transferred is given by:
Q = mcΔT
where Q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature.
Given:
Mass of lead (m_lead) = 0.400 kg
Mass of tin (m_tin) = 0.400 kg
Mass of water (m_water) = 1.00 kg
Initial temperature of lead and tin (T_lead and T_tin) = 60.0°C
Initial temperature of water (T_water) = 20.0°C
First, let's calculate the heat transferred from the lead:
Q_lead = m_lead * c_lead * ΔT_lead
For lead, the specific heat (c_lead) is given as 0.13 kJ/kg°C. The change in temperature (ΔT_lead) is given by:
ΔT_lead = T_final - T_lead
where T_final is the final temperature of the system. Since we're assuming the alloy is at equilibrium, T_final will be the same for both lead and tin.
Next, let's calculate the heat transferred from the tin:
Q_tin = m_tin * c_tin * ΔT_tin
For tin, the specific heat (c_tin) is given as 0.22 kJ/kg°C. The change in temperature (ΔT_tin) is the same as ΔT_lead.
Now, let's calculate the heat gained by the water:
Q_water = m_water * c_water * ΔT_water
For water, the specific heat (c_water) is given as 4.18 kJ/kg°C. The change in temperature (ΔT_water) is the same as ΔT_lead and ΔT_tin.
Since the heat lost by the lead and tin is equal to the heat gained by the water, we can write:
Q_lead + Q_tin = Q_water
m_lead * c_lead * ΔT + m_tin * c_tin * ΔT = m_water * c_water * ΔT
Simplifying, we have:
(m_lead * c_lead + m_tin * c_tin) * ΔT = m_water * c_water * ΔT
(m_lead * c_lead + m_tin * c_tin) = m_water * c_water
To find the specific heat of the alloy (c_alloy), we divide both sides by the total mass of the alloy:
c_alloy = (m_lead * c_lead + m_tin * c_tin) / (m_lead + m_tin)
Substituting the given values:
c_alloy = (0.400 kg * 0.13 kJ/kg°C + 0.400 kg * 0.22 kJ/kg°C) / (0.400 kg + 0.400 kg)
Simplifying, we get:
c_alloy = (0.052 kJ + 0.088 kJ) / 0.800 kg
c_alloy = 0.140 kJ / 0.800 kg
c_alloy = 0.175 kJ/kg°C
Therefore, the specific heat anticipated for the alloy is 0.175 kJ/kg°C.