Question

What mass of steam at \(100^{\circ} \mathrm{C}\) must be mixed with \(150 \mathrm{~g}\) of ice at \(0^{\circ} \mathrm{C}\), in a thermally insulated container, to produce liquid water at \(50^{\circ} \mathrm{C}\) ?

Short answer

The result of the calculation in step 5 will yield the mass of the steam required. Note that as the calculation involves energy being conserved, the result is dependent on the specific values of the system.

Step-by-step solution

Define the total energy exchange

The total energy exchange can be defined as the sum of the energy used to melt the ice, heat the water from \(0^{\circ}C\) to \(50^{\circ}C\), and condense the steam to water then cool it to \(50^{\circ}C\). Thus, the equation can be written as: \[Q_{total}=Q_{ice->water} + Q_{water-heating} + Q_{steam->water} + Q_{cooling-steam}\]

Calculate the energy to melt the ice

The energy required to convert ice at \(0^{\circ}C\) to water at \(0^{\circ}C\) can be calculated using the formula: \[Q_{ice->water} = m_{ice} * L_{f}\] where \(m_{ice} = 150g\) is the mass of the ice and \(L_{f} = 334 J/g\) is the latent heat of fusion of ice.

Calculate the energy to heat the water from \(0^{\circ}C\) to \(50^{\circ}C\)

The energy needed to heat the resulting water from \(0^{\circ}C\) to \(50^{\circ}C\) can be calculated using the formula: \[Q_{water-heating} = m_{water} * c_{water} * \Delta T\] where \(m_{water} = 150g\) is the mass of the resulting water, \(c_{water} = 4.18 J/g.\degree C\) is the specific heat capacity of water, and \(\Delta T = 50^{\circ}C - 0^{\circ}C = 50^{\circ}C\) is the change in temperature.

Calculate the energy to convert steam to water then cool it to \(50^{\circ}C\)

The energy taken by the steam to condense to water at \(100^{\circ}C\) and then cool down to \(50^{\circ}C\) is calculated with the formula: \[Q_{steam->water} = m_{steam} * L_v\] and \[Q_{cooling-steam} = m_{steam} * c_{water} * \Delta T\] respectively. Here, \(m_{steam}\) is the mass of steam we are asked to find, \(L_v = 2260 J/g\) is the latent heat of vaporization of water, \(c_{water}\) is the specific heat capacity of water, and \(\Delta T = 100^{\circ}C - 50^{\circ}C = 50^{\circ}C\) is the change in temperature.

Solve for the mass of the steam

Setting the energy given by the steam to be equal to the energy taken in by the ice and the resulting water, the mass of the steam, \(m_{steam}\), is found by solving the equation: \[Q_{total}=0\] This will lead to the value of the mass of the steam needed.

Key concepts

Latent Heat of Fusion

The latent heat of fusion is an essential concept in thermodynamics. It refers to the energy required to change a substance from its solid to liquid state without changing its temperature. For ice, this value is particularly crucial when considering the conversion of ice at \(0^{\circ}C\) to water at the same temperature.

This process involves breaking the molecular bonds within ice, which requires energy. The latent heat of fusion for ice is \(334 \text{ J/g}\). Therefore, if you have 150 grams of ice, the energy needed to melt it completely into water at \(0^{\circ}C\) is calculated using the formula:

• \( Q_{ice \to water} = m_{ice} \times L_{f} \)
• \( m_{ice} = 150\, \text{g} \)
• \( L_{f} = 334\, \text{J/g} \)

Thus, understanding the latent heat of fusion allows for the accurate calculation of energy changes during the phase transition from solid ice to liquid water.

Latent Heat of Vaporization

The latent heat of vaporization involves the energy needed to change a substance from a liquid to a gas at a constant temperature. In this scenario, it particularly refers to the energy changes when steam condenses back into water.

For water, the latent heat of vaporization is relatively high at \(2260 \text{ J/g}\). This is because converting a liquid to a gas requires more energy to overcome the intermolecular attractions. Understanding this concept helps in problems where steam transforms into liquid water, especially when calculating energy exchanges.

When steam at \(100^{\circ}C\) becomes water, the formula used is:

• \( Q_{steam \to water} = m_{steam} \times L_{v} \)
• \( L_{v} = 2260\, \text{J/g} \)

This relationship is fundamental in assessing how much energy is involved in the phase transformation of water vapor to liquid.

Specific Heat Capacity

The specific heat capacity indicates how much energy is required to raise the temperature of a unit mass of a substance by 1 degree Celsius. This property is crucial in the study of thermal interactions and energy exchange within substances.

For water, the specific heat capacity is \(4.18 \text{ J/g.}^{\circ}C\). This means that warming up water from \(0^{\circ}C\) to \(50^{\circ}C\) requires a significant amount of energy since water can store heat efficiently.

In equations, it is represented as:

• \( Q = m \times c \times \Delta T \)
• \( m \) is the mass of the substance in grams
• \( c \) is the specific heat capacity
• \( \Delta T \) is the change in temperature

This concept allows us to predict how substances will behave under temperature changes and is central to thermal equilibrium problems.

Thermal Equilibrium

Thermal equilibrium in thermodynamics refers to a state where all parts of a system reach the same temperature and there is no net exchange of energy between them.

This principle is key to solving problems involving energy transfer, like mixing steam and ice. The total energy added to a system equals the total energy lost by the system, leading to a state where input and output energy are balanced. This is expressed in the equation for the exercise:

• \( Q_{total} = 0 \)

When ice and steam are mixed in a closed system, without external energy inputs or losses, thermal equilibrium calculates the conditions under which the final temperature is reached.

Understanding thermal equilibrium helps explain and predict how temperature adjustments occur naturally, promoting accurate assessments in heat transfer situations.