A piece of lead at 100 degrees celsius has a mass of 500 grams. It is placed in a 200 grams of water at 20 degrees celsius that is contained in an aluminum caloriemeter with a mass of 400 grams. If the final temperature of lead, water and caloriemeter is 24 degrees celcius, what is the specific heat of the lead
To find the specific heat of lead, we can use the principle of energy conservation or the heat transfer equation.
The principle of energy conservation states that the total heat gained by the lead, water, and caloriemeter equals the total heat lost by the lead:
Heat gained = Heat lost
The heat gained by the water and caloriemeter can be calculated using the formula:
Q = mcΔT
Where:
Q is the heat gained by the water and caloriemeter,
m is the sum of the masses of the water and caloriemeter,
c is the specific heat capacity (constant) of water, which is 4.18 J/g°C (joules per gram-degree Celsius), and
ΔT is the change in temperature of the water and caloriemeter (final temperature - initial temperature).
First, let's calculate the heat gained by the water and caloriemeter:
Q_water_caloriemeter = (m_water + m_caloriemeter) * c_water * ΔT_water_caloriemeter
Where:
m_water = mass of water = 200 g
m_caloriemeter = mass of caloriemeter = 400 g
c_water = specific heat capacity of water = 4.18 J/g°C
ΔT_water_caloriemeter = change in temperature of water and caloriemeter = (final temperature - initial temperature) = (24°C - 20°C) = 4°C
Plugging in the values:
Q_water_caloriemeter = (200 g + 400 g) * 4.18 J/g°C * 4°C
Q_water_caloriemeter = 2400 J
Now, since the heat gained by the water and caloriemeter is equal to the heat lost by the lead, we can use the formula:
Q_lead = m_lead * c_lead * ΔT_lead
Where:
Q_lead = heat lost by the lead
m_lead = mass of lead = 500 g
c_lead = specific heat capacity of lead (what we want to find)
ΔT_lead = change in temperature of lead = (final temperature - initial temperature) = (24°C - 100°C) = -76°C (negative because the lead is losing heat)
Plugging in the values and rearranging the equation to solve for c_lead:
Q_lead = m_lead * c_lead * ΔT_lead
c_lead = Q_lead / (m_lead * ΔT_lead)
c_lead = 2400 J / (500 g * -76°C)
c_lead = -2400 J / (500 g * 76°C)
c_lead ≈ -0.063 J/g°C
The specific heat of lead is approximately -0.063 J/g°C. Note that the negative sign indicates that lead loses heat when heated instead of gaining it like water.