Question

An ice cube with a mass of 20 \(\mathrm{g}\) at \(-20^{\circ} \mathrm{C}\) (typical freezer temperature) is dropped into a cup that holds 500 \(\mathrm{mL}\) of hot water, initially at \(83^{\circ} \mathrm{C} .\) What is the final temperature in the cup? The density of liquid water is 1.00 \(\mathrm{g} / \mathrm{mL}\) ; the specific heat capacity of ice is \(2.03 \mathrm{J} / \mathrm{g}-\mathrm{C}\) ; the specific heat capacity of liquid water is \(4.184 \mathrm{J} / \mathrm{g}-\mathrm{C} ;\) the enthalpy of fusion of water is 6.01 \(\mathrm{k} \mathrm{J} / \mathrm{mol} .\)

Short answer

The final temperature in the cup after adding the ice cube is approximately \(48.9^{\circ} \mathrm{C}\).

Step-by-step solution

Identify energy conservation equation

The equation for energy conservation in this case is: Energy gained by ice + energy for phase change = energy lost by hot water

Write expressions for energy gained, energy for phase change, and energy lost

- Energy gained by ice to reach the melting point \(q_1\): \(q_1 = m_i c_i (0 - (-20))\) - Energy needed for the phase change \(q_2\): \(q_2 = m_i H_f\) - Energy gained by ice to reach the final temperature \(q_3\): \(q_3 = m_i c_w (T_f - 0)\) - Hot water energy lost \(q_4\): \(q_4 = m_w c_w (T_f - 83)\) Where \(m_i\) is the mass of ice, \(c_i\) is the specific heat capacity of ice, \(m_w\) is the mass of hot water, and \(c_w\) is the specific heat capacity of liquid water. H_f is the enthalpy of fusion, and T_f is the final temperature, which is the same for the ice and water mixture.

Substitute given values

- \(m_i = 20 \: g\) - \(c_i = 2.03 \: \frac{J}{g \cdot C}\) - \(m_w = 500 \: mL = 500 \: g\) (since the density of liquid water is 1.00 g/mL) - \(c_w = 4.184 \: \frac{J}{g \cdot C}\) - \(H_f = 6.01 \: \frac{kJ}{mol} = 6.01 \times 10^3 \: \frac{J}{mol}\)

Calculate the number of moles of ice

Molecular weight of water = 18 g/mol Number of moles of ice \(n_i\) = \(\frac{20 \: g}{18 \: \frac{g}{mol}} = 1.11 \: mol\)

Apply energy conservation equation and solve for T_f

Using the energy conservation equation, and substituting the given values, we obtain: 20 * 2.03 * 20 + 1.11 * (6.01 * 10^3) + 20 * 4.184 * T_f = 500 * 4.184 * (T_f - 83) Now, solve for T_f. This requires simplifying and solving the equation.

Simplify and solve the equation

Upon simplification and re-arrangement, we get the equation: \(20*2.03*20 + 1.11*(6.01*10^3) - 500 * 4.184 * 83 = (500*4.184 - 20*4.184) * T_{f}\) Now solve for T_f: \(T_f = \frac{20*2.03*20 + 1.11*(6.01*10^3) - 500 * 4.184 * 83}{500*4.184 - 20*4.184}\) \(T_f \approx 48.9^{\circ} \mathrm{C}\)

Final result

The final temperature in the cup after adding the ice cube is approximately \(48.9^{\circ} \mathrm{C}\).

Key concepts

Energy Conservation

In the context of thermodynamics, energy conservation is a fundamental concept asserting that energy cannot be created or destroyed; rather, it just changes form. Applied to chemistry and physics, this principle allows us to understand how energy is transferred between substances during chemical reactions and physical processes.

For example, when an ice cube melts in hot water, the water loses some of its thermal energy to the ice, which gains that energy, causing the ice to warm up and eventually turn into liquid. This process is beautifully illustrated by the energy conservation equation used in the given exercise:

• The energy gained by the ice as it warms up to the melting point,
• Plus the energy needed for the phase change from ice to liquid water,
• Is equal to the energy lost by the hot water.

By using this approach, we can track the energy flow and predict the final temperature of the water.

Specific Heat Capacity

The specific heat capacity of a substance is a measure of how much energy is required to raise the temperature of one gram of the substance by one degree Celsius (or Kelvin). It's an intrinsic property that varies from material to material and plays a pivotal role in determining how substances absorb and transfer heat.

In our exercise, two different specific heat capacities are considered: one for ice (c_i = 2.03 J/g·°C) and another for liquid water (c_w = 4.184 J/g·°C). The varying values imply that it takes more energy to heat water than to heat ice by the same temperature increment. Intuitively, this means ice will warm up faster when energy is added as compared to water, a concept crucial for solving our textbook problem.

Enthalpy of Fusion

The enthalpy of fusion, also known as the heat of fusion, refers to the energy required to change a substance from the solid phase to the liquid phase at its melting point without changing its temperature. It's an essential factor when calculating the energy involved in a phase change, as seen in the melting of ice in our example.

Water's relatively high enthalpy of fusion (H_f = 6.01 kJ/mol) indicates that melting ice demands a considerable amount of energy. This is because breaking the hydrogen bonds in ice, which keep water molecules in a fixed structure, consumes a lot of energy. In our exercise, we require the enthalpy of fusion to calculate the energy needed for the ice to transition to liquid, one of the critical steps to finding the final temperature of the system.

Phase Change

A phase change is a transition of matter from one state to another, such as from solid to liquid (melting), liquid to gas (boiling), or solid to gas (sublimation). During a phase change, the temperature of a substance doesn’t increase even though energy is being absorbed (or released), until the transition is complete.

In our hot water and ice example, understanding phase changes is vital. The ice cube undergoes a phase change from solid to liquid before it can reach thermal equilibrium with the hot water. During this process, the temperature of the ice doesn't increase while melting (even though it is absorbing heat), which reflects the importance of incorporating the enthalpy of fusion into our calculations for a proper understanding and accurate result.