Question

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 \(6.00 \mathrm{~mL}\) of water in a Pyrex glass vial thermally insulated from the environment. The empty vial has a mass of \(5.00 \mathrm{~g}\). The thermometer you use is made of Pyrex glass as well and has a mass of \(15.0 \mathrm{~g}\), of which \(4.00 \mathrm{~g}\) is the mercury inside the thermometer. The thermometer is initially at room temperature \(\left(20.0^{\circ} \mathrm{C}\right) .\) You place the thermometer in the water in the vial and, after a while, you read an equilibrium temperature of \(29.0^{\circ} \mathrm{C} .\) 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 \(800 . J /(\mathrm{kg} \mathrm{K})\) and that of liquid mercury at room temperature is \(140 . \mathrm{J} /(\mathrm{kg} \mathrm{K})\)

Short answer

Solution: We are given the following values: - Mass of water: \(m_{w} = 6.00 \times 10^{-3} \mathrm{kg}\) - Mass of vial: \(m_{v} = 5.00 \times 10^{-3} \mathrm{kg}\) - Mass of glass thermometer: \(m_{t} = 15.0 \times 10^{-3} \mathrm{kg}\) - Mass of mercury inside thermometer: \(m_{m} = 4.00 \times 10^{-3} \mathrm{kg}\) - Initial temperature: \(T_{i} = 20.0^{\circ} \mathrm{C}\) - Equilibrium temperature: \(T_{f} = 29.0^{\circ} \mathrm{C}\) - Specific heat capacity of Pyrex glass: \(c_{g} = 800 . J /(\mathrm{kg} \mathrm{K})\) - Specific heat capacity of liquid mercury: \(c_{m} = 140 . \mathrm{J} / (\mathrm{kg} \mathrm{K})\) - Specific heat capacity of water: \(c_{w} = 4.18 \times 10^3 . \mathrm{J} / (\mathrm{kg} \mathrm{K})\) \(Q_{g} = (5.00 \times 10^{-3} + 15.0 \times 10^{-3})(800)(9.0) = 0.144 \cdot 10^3 J\) \(Q_{m} = (4.00 \times 10^{-3})(140)(9.0) = 5.040 \cdot 10^0 J\) \(Q_{total} = Q_{g} + Q_{m} = 0.144 \cdot 10^3 + 5.040 \cdot 10^0 = 149.04 J\) \(T_{w_{i}} = T_{f} - \dfrac{Q_{total}}{m_{w} c_{w}} = 29.0 - \dfrac{149.04}{(6.00 \times 10^{-3})(4.18 \times 10^3)} \) \(T_{w_{i}} ≈ 32.4^{\circ} \mathrm{C}\) The actual temperature of the water in the vial before the temperature was measured was approximately \(32.4^{\circ} \mathrm{C}\).

Step-by-step solution

List known values

We are given the following values: - Mass of water: \(m_{w} = 6.00 \mathrm{~mL} = 6.00 \mathrm{~g}\) (1 mL of water weighs approximately 1 g) - Mass of vial: \(m_{v} = 5.00 \mathrm{~g}\) - Mass of glass thermometer: \(m_{t} = 15.0 \mathrm{~g}\) - Mass of mercury inside thermometer: \(m_{m} = 4.00 \mathrm{~g}\) - Initial temperature: \(T_{i} = 20.0^{\circ} \mathrm{C}\) - Equilibrium temperature: \(T_{f} = 29.0^{\circ} \mathrm{C}\) - Specific heat capacity of Pyrex glass: \(c_{g} = 800 . J /(\mathrm{kg} \mathrm{K})\) - Specific heat capacity of liquid mercury: \(c_{m} = 140 . \mathrm{J} / (\mathrm{kg} \mathrm{K})\) - Specific heat capacity of water: \(c_{w} = 4.18 \times 10^3 . \mathrm{J} / (\mathrm{kg} \mathrm{K})\) Note that we will need to convert the masses to kg to match the units of specific heat capacities.

Convert masses to kilograms

Convert the given mass values to kilograms: - \(m_{w} = 6.00 \times 10^{-3} \mathrm{~kg}\) - \(m_{v} = 5.00 \times 10^{-3} \mathrm{~kg}\) - \(m_{t} = 15.0 \times 10^{-3} \mathrm{~kg}\) - \(m_{m} = 4.00 \times 10^{-3} \mathrm{~kg}\)

Calculate heat gained by the vial and thermometer

The heat gained by the vial and thermometer (assuming only the glass part gains heat) can be calculated using the equation: \(Q = mc\Delta T\) For glass parts of the vial and thermometer: - \(\Delta T_{g} = T_{f} - T_{i} = 29.0 - 20.0 = 9.0^{\circ} \mathrm{C}\) \(Q_{g} = (m_{v} + m_{t})c_{g}\Delta T_{g}\) \(Q_{g} = (5.00 \times 10^{-3} + 15.0 \times 10^{-3})(800)(9.0)\) For the mercury inside the thermometer: - \(\Delta T_{m} = T_{f} - T_{i} = 9.0^{\circ} \mathrm{C}\) \(Q_{m} = m_{m} c_{m} \Delta T_{m}\) \(Q_{m} = (4.00 \times 10^{-3})(140)(9.0)\) Now, calculate the total heat gained as \(Q_{total} = Q_{g} + Q_{m}\)

Calculate heat lost by the water and find initial temperature

Since the heat gained by the thermometer and vial equals the heat lost by the water: \(Q_{total} = m_{w} c_{w} (T_{f} - T_{w_{i}})\), where \(T_{w_{i}}\) is the initial temperature of the water. Solve for \(T_{w_{i}}\): \(T_{w_{i}} = T_{f} - \dfrac{Q_{total}}{m_{w} c_{w}} \) Plug in the values and calculate \(T_{w_{i}}\).

Interpret the result

After calculating the initial temperature of the water (\(T_{w_{i}}\)), interpret the result and state the actual temperature of the water in the vial before the temperature was measured.

Key concepts

Specific Heat Capacity

Specific heat capacity is a property that tells us how much heat energy is required to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin). It’s like a measure of how much heat 'patience' a material has – the higher the specific heat capacity, the more heat it can absorb without getting significantly hotter. For example, water has a high specific heat capacity, which is why it takes a long time to boil.

In the context of our thermometer and water vial problem, we need to know the specific heat capacities of Pyrex glass and mercury to understand how they affect the measurement of the water’s temperature. When we place a thermometer in water, both the water and the thermometer exchange heat until they reach an equilibrium temperature. Due to this exchange, the thermometer absorbs some of the water's heat and hence, the temperature we read (equilibrium) is not the true initial temperature of the water.

Heat Transfer

Heat transfer is the process of thermal energy being exchanged between different materials or areas. In our scenario, with the thermometer placed in the water, we’re witnessing heat transfer from the warmer water to the cooler thermometer until they reach the same temperature.

This transferral of energy occurs until the system reaches thermal equilibrium – a state where all parts of the system are at the same temperature, meaning there is no further heat flow between them. The energy transferred is called heat and is measured in joules (J) in the metric system. The amount of heat transferred depends on the mass, the change in temperature, and the specific heat capacity of the substances involved, which leads us directly to the next concept, equilibrium temperature.

Equilibrium Temperature

The equilibrium temperature is the uniform temperature of a system when all parts have the same temperature, and heat transfer has stabilized. It's like when two friends with different opinions talk it out and reach a mutual conclusion. Neither the water nor the thermometer will change temperature once they've reached this point, assuming there is no heat loss to the surroundings.

In our textbook exercise, the immersion of the thermometer in water results in the transfer of heat between the water and the thermometer leading to an equilibrium temperature, which is different from the water's initial temperature. By taking into account the specific heat capacities of the materials involved and the heat gained or lost, we can backtrack to find out the true initial temperature of the water before the introduction of the thermometer.

Thermal Insulation

Thermal insulation refers to materials or methods used to reduce the rate of heat transfer. Just think of it as how a cozy blanket slows down the escape of your body heat on a cold night. In our example, the Pyrex glass vial is thermally insulated, meaning it minimizes heat loss or gain with the environment.

Effective insulation is crucial in our setup because it ensures that the heat exchange occurs only between the thermometer and the water, without interference from outside temperatures. This isolation is essential for calculating the actual initial temperature of the water since, without it, additional variables would make the problem more complex and the solution less accurate.