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

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

Step by step solution

01

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
02

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.
03

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) - \(c_w = 4.184 \: \frac{J}{g \cdot C}\) - \(H_f = 6.01 \: \frac{kJ}{mol} = 6.01 \times 10^3 \: \frac{J}{mol}\)
04

Calculate the number of moles of ice

Molecular weight of water = 18 g/mol Number of moles of ice \(n_i\) = \(\frac{20 \: g}{18 \: \frac{g}{mol}} = 1.11 \: mol\)
05

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.
06

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: \(T_f = \frac{20*2.03*20 + 1.11*(6.01*10^3) - 500 * 4.184 * 83}{500*4.184 - 20*4.184}\) \(T_f \approx 48.9^{\circ} \mathrm{C}\)
07

Final result

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 (ci = 2.03 J/g·°C) and another for liquid water (cw = 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 (Hf = 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.

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Most popular questions from this chapter

Indicate whether each statement is true or false. (a) The entropy of the universe increases for any spontaneous process. (b) The entropy change of the system is equal and opposite that of the surroundings for any irreversible process. (c) The entropy of the system must increase in any spontaneous process. (a) The entropy change for an isothermal process depends on both the absolute temperature and the amount of heat reversibly transferred.

The reaction $$ \mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{S}(g) \rightleftharpoons 3 \mathrm{S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ is the basis of a suggested method for removal of SO \(_{2}\) from power-plant stack gases. The standard free energy of each substance is given in Appendix C. (a) What is the equilibrium constant for the reaction at 298 \(\mathrm{K} ?\) (b) In principle, is this reaction a feasible method of removing \(S O_{2} ?\) (c) If \(P_{S O_{2}}=P_{H_{2} S}\) and the vapor pressure of water is 25 torr, calculate the equilibrium \(\mathrm{SO}_{2}\) pressure in the system at 298 K. (d) Would you expect the process to be more or less effective at higher temperatures?

When most elastomeric polymers (e.g., a rubber band) are stretched, the molecules become more ordered, as illustrated here: Suppose you stretch a rubber band. (a) Do you expect the entropy of the system to increase or decrease? (b) If the rubber band were stretched isothermally, would heat need to be absorbed or emitted to maintain constant temperature? (c) Try this experiment: Stretch a rubber band and wait a moment. Then place the stretched rubber band on your upper lip, and let it return suddenly to its unstretched state (remember to keep holding on!). What do you observe? Are your observations consistent with your answer to part (b)?

In chemical kinetics, the entropy of activation is the entropy change for the process in which the reactants reach the activated complex. Predict whether the entropy of activation for a bimolecular process is usually positive or negative.

The normal boiling point of \(\mathrm{Br}_{2}(l)\) is \(58.8^{\circ} \mathrm{C},\) and its molar enthalpy of vaporization is \(\Delta H_{\mathrm{vap}}=29.6 \mathrm{kJ} / \mathrm{mol}\) (a) When \(\mathrm{Br}_{2}(l)\) boils at its normal boiling point, does its entropy increase or decrease? (b) Calculate the value of \(\Delta S\) when 1.00 mol of \(\mathrm{Br}_{2}(l)\) is vaporized at \(58.8^{\circ} \mathrm{C}\) .

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