When an immersion glass thermometer is used to measure the temperature of a liquid, the temperature reading will be affected by an error due to heat transfer between the liquid and the thermometer. Suppose you want to measure the temperature of 7 mL of water in a Pyrex glass vial thermally insulated from the environment. The empty vial has a mass of 5.0 g. The thermometer you use is made of Pyrex glass as well and has a mass of 18 g, of which 4.5 g is the mercury inside the thermometer. The thermometer is initially at room temperature (20.0°C). You place the thermometer in the water in the vial and, after a while, you read an equilibrium temperature of 29°C. What was the actual temperature of the water in the vial before the temperature was measured? The specific heat capacity of Pyrex glass around room temperature is 800 J/(kg K) and that of liquid mercury at room temperature is 140 J/(kg K).
-Qout=Qin
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-(Qwater+Qvial)=Qthermometer+Qmercury
-(mwater*cwater*(Tfinal-Tinitial)+mvial*cvial*(Tfinal-Tinitial)) = mthermometer*cthermometer*(Tfinal-Tinitial)+mmercury*cmercury*(Tfinal-Tinitial)
Plug in the equilibrium temperature for Tfinal. Then plug in the Room temperature for Thermometer and Mercury Tinitial. Solve for Water and Vial Tinitial. Done.
Make sure to use the right C numbers if you change the masses to kg.
To determine the actual temperature of the water in the vial before the temperature measurement, we need to consider the heat transfer between the water, the vial, and the thermometer.
First, let's calculate the heat transfer from the water to the thermometer and vial.
The heat transferred from the water can be calculated using the equation:
Q_water = m_water * c_water * ΔT
Where:
Q_water is the heat transferred from the water
m_water is the mass of the water
c_water is the specific heat capacity of water
ΔT is the change in temperature of the water
Given data:
m_water = 7 g
c_water = 4184 J/(kg K) (specific heat capacity of water)
ΔT = 29°C - 20°C = 9 K
Substituting the values:
Q_water = 7 g * 4184 J/(kg K) * 9 K
Q_water = 265848 J
Next, let's calculate the heat transferred to the vial.
The heat transferred from the water to the vial can be calculated using the equation:
Q_vial = m_vial * c_vial * ΔT
Where:
Q_vial is the heat transferred from the water to the vial
m_vial is the mass of the vial
c_vial is the specific heat capacity of Pyrex glass
Given data:
m_vial = 5 g
c_vial = 800 J/(kg K)
ΔT = 9 K (same as before)
Substituting the values:
Q_vial = 5 g * 800 J/(kg K) * 9 K
Q_vial = 36000 J
Now, let's consider the heat transferred to the thermometer.
The heat transferred from the water to the thermometer can be calculated using the equation:
Q_thermometer = m_thermometer * c_thermometer * ΔT
Where:
Q_thermometer is the heat transferred from the water to the thermometer
m_thermometer is the total mass of the thermometer (including the mercury inside it)
c_thermometer is the specific heat capacity of Pyrex glass (same as the vial)
Given data:
m_thermometer = 18 g
c_thermometer = 800 J/(kg K)
ΔT = 9 K (same as before)
Substituting the values:
Q_thermometer = 18 g * 800 J/(kg K) * 9 K
Q_thermometer = 129600 J
Now, let's calculate the heat transferred from the thermometer to the surroundings.
The heat transferred from the thermometer to the surroundings (the room) is given by:
Q_surroundings = m_thermometer * c_thermometer * ΔT
Where:
Q_surroundings is the heat transferred from the thermometer to the surroundings (room temperature)
m_thermometer is the total mass of the thermometer (including the mercury inside it)
c_thermometer is the specific heat capacity of Pyrex glass (same as the vial)
ΔT is the change in temperature of the thermometer, which is the difference between the equilibrium temperature and the initial temperature.
Given data:
m_thermometer = 18 g
c_thermometer = 800 J/(kg K)
ΔT = 29°C - 20°C = 9 K
Substituting the values:
Q_surroundings = 18 g * 800 J/(kg K) * 9 K
Q_surroundings = 129600 J
Now, let's consider the heat absorbed by the mercury inside the thermometer.
The heat absorbed by the mercury can be calculated using the equation:
Q_mercury = m_mercury * c_mercury * ΔT
Where:
Q_mercury is the heat absorbed by the mercury
m_mercury is the mass of the mercury
c_mercury is the specific heat capacity of mercury
ΔT is the change in temperature of the mercury
Given data:
m_mercury = 4.5 g
c_mercury = 140 J/(kg K)
ΔT = 9 K (same as before)
Substituting the values:
Q_mercury = 4.5 g * 140 J/(kg K) * 9 K
Q_mercury = 5670 J
Now, let's consider the heat lost by the vial and the thermometer to the surroundings.
The heat lost by the vial and the thermometer to the surroundings can be calculated as the sum of the heat transferred to the surroundings from the vial and the thermometer:
Q_loss = Q_vial + Q_thermometer + Q_surroundings
Substituting the values:
Q_loss = 36000 J + 129600 J + 129600 J
Q_loss = 295200 J
Now, let's determine the heat transferred from the water to the thermometer:
Q_transfer = Q_water - Q_loss
Substituting the values:
Q_transfer = 265848 J - 295200 J
Q_transfer = -29352 J
Since the heat transfer is negative, it implies that heat was transferred from the thermometer to the water.
Now, let's calculate the change in temperature of the thermometer due to this heat transfer:
ΔT_thermometer = Q_transfer / (m_thermometer * c_thermometer)
Substituting the values:
ΔT_thermometer = -29352 J / (18 g * 800 J/(kg K))
ΔT_thermometer = -1.022 K
Since the temperature of the thermometer is initially at 20°C (293K), the actual temperature of the water in the vial before the temperature measurement is:
Actual temperature = Equilibrium temperature - ΔT_thermometer
Actual temperature = 29°C - (-1.022 K) = 30.022°C (or approximately 30.0°C)
Therefore, the actual temperature of the water in the vial before the temperature measurement was approximately 30.0°C.