Question

A 125 \(g\) stainless steel ball bearing \((\mathrm{spht}=\) \(0.50 \mathrm{Jg}^{-1}\) \(\left.^{\circ} \mathrm{C}^{-1}\right)\) at \(525^{\circ} \mathrm{C}\) is dropped into \(75.0 \mathrm{mL}\) of water at \(28.5^{\circ} \mathrm{C}\) in an open Styrofoam cup. As a result, the water is brought to a boil when the temperature reaches \(100.0^{\circ} \mathrm{C} .\) What mass of water vaporizes while the boiling continues? \(\left(\Delta H_{\mathrm{vap}}^{\circ}=40.6 \mathrm{kJ} / \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\right)\).

Short answer

The mass of the water that vaporizes can be found by using the law of conservation of energy and the heat of vaporization of water.

Step-by-step solution

Calculate heat lost by the steel ball.

The first step is to calculate the heat (q) lost by the steel ball bearing as it cools from \(525^{\circ} C\) to \(100^{\circ} C\). The formula to use is \(q_{\text{steel}} = m_{\text{steel}} \cdot c_{\text{steel}} \cdot \Delta T_{\text{steel}}\), where \(m_{\text{steel}} = 125 \, g\) is the mass of the steel, \(c_{\text{steel}} = 0.50 \, J \, g^{-1} \, ^{\circ}C^{-1}\) is the specific heat of the steel, and \(\Delta T_{\text{steel}}\) is the change in temperature of the steel, which is \(525^{\circ} C - 100^{\circ} C = 425^{\circ} C\).

Calculate initial heat absorbed by water to reach boiling point.

Next, calculate the heat (q) absorbed by the water as it heats up from \(28.5^{\circ} C\) to \(100^{\circ} C\). The formula to use is \(q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}}\), where \(m_{\text{water}} = 75.0 \, g\) (since 1mL of water has mass of 1g), \(c_{\text{water}} = 4.18 \, J \, g^{-1} \, ^{\circ}C^{-1}\), and \(\Delta T_{\text{water}}\) is the change in temperature of the water, which is \(100.0^{\circ} C - 28.5^{\circ} C = 71.5^{\circ} C\).

Find heat absorbed by water to vaporize.

The amount of heat (q) required to vaporize the water can be found from the difference of the heat lost by the steel and the heat initially absorbed by the water to reach the boiling point, i.e., \(q_{\text{vaporize}} = q_{\text{steel}} - q_{\text{water}}\).

Calculate mass of water vaporized.

Finally, find the mass of the water that vaporizes. This can be found by using the formula \(q_{\text{vaporize}} = m_{\text{vaporized}} \cdot \Delta H_{\text{vap}}\), where \(\Delta H_{\text{vap}} = 40.6 \, kJ/mol\) is the heat of vaporization of the water. The mass can be found by isolating it in the above formula: \(m_{\text{vaporized}} = \frac{q_{\text{vaporize}}}{\Delta H_{\text{vap}}}\). Note that \(\Delta H_{\text{vap}}\) needs to be converted to \(J \, g^{-1}\) to match the units with \(q_{\text{vaporize}}\).

Key concepts

Heat Transfer

When an object at a high temperature comes into contact with another object at a lower temperature, heat transfer occurs. This principle is key in the exercise you are working on. The stainless steel ball bearing starts at a high temperature of \(525^{\circ}C\) and is placed into water at \(28.5^{\circ}C\). Due to the difference in temperature, heat moves from the steel to the water.

Heat transfer is a natural process aimed at reaching thermal equilibrium, meaning both substances eventually have the same temperature. The amount of heat transferred depends on the specific heat capacity of the materials, which helps determine how much heat is required to change the temperature of a given mass of material by a degree Celsius. In this problem, heat is transferred until the water begins to boil.

Specific Heat Capacity

Specific heat capacity (\(c\)) is a material's ability to absorb heat. It tells us how much energy is needed to raise the temperature of 1 gram of the material by 1 degree Celsius. This concept plays a vital role when calculating the heat lost or gained by substances.

For example, the steel ball bearing has a specific heat capacity of \(0.50 \, \text{J} \, \text{g}^{-1} \, ^{\circ}\text{C}^{-1}\). This informs us that each gram of the steel requires 0.50 Joules to increase its temperature by 1 degree Celsius. Similarly, water has a specific heat capacity of \(4.18 \, \text{J} \, \text{g}^{-1} \ ^{\circ}\text{C}^{-1}\), which is why water is excellent at retaining heat.

• Higher specific heat means more energy required to change temperature.
• In contrast, lower specific heat indicates less energy needed.

Knowing the specific heat helps calculate how much heat energy (\(q\)) is involved as the temperature changes, helping solve for the energy balance in such problems.

Heat of Vaporization

The heat of vaporization is the amount of energy required to convert a liquid into vapor without a temperature change. It is expressed as energy per mole of substance, but in this problem, we convert it to energy per gram. For water, the heat of vaporization is \(\Delta H_{\text{vap}} = 40.6 \, \text{kJ/mol}\), which is needed for water to change from liquid at boiling point to vapor.

When solving, you find how much of this energy is needed when the water reaches \(100^{\circ}C\) and starts vaporizing. The excess heat from the steel is used to vaporize some water. To find how much water vaporizes, the energy from the heat transfer is divided by the heat of vaporization, making it easier to see how molecules transition from liquid to gas.

Energy Conservation

In thermochemistry, energy conservation is a fundamental idea that states energy cannot be created or destroyed, only transferred or converted. This principle is practiced in calorimetry problems like this one to solve for unknown variables.

In the exercise, the system follows energy conservation where energy released by the steel ball (as it cools) is equal to the energy gained by water (to heat and vaporize). This balance allows you to find missing values, such as the mass of water vaporized. Here's how the process works:

• Calculate heat lost by the hot object (steel).
• Measure heat gained by the other substance (water) to reach a new state.
• The remaining energy helps determine the extent of phase changes like vaporization.

Understanding energy conservation ensures that all energy changes in interactions are accounted for, leading to accurate and reliable results.