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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 \(6.01 \mathrm{~kJ} / \mathrm{mol}\).

Short Answer

Expert verified
The final temperature in the cup is approximately \(79.42^{\circ}\mathrm{C}\).

Step by step solution

01

Calculate the energy required to raise ice to 0°C

First, we need to find the energy required to raise the temperature of the ice from -20°C to 0°C. We will use the formula Q = mcΔT, with m = 20g, c = 2.03 J/g°C, and ΔT = 20°C. \(Q_{1} = mc\Delta T\) \(Q_1 = (20 \mathrm{~g})(2.03 \mathrm{~J/g°C})(20°C)\) \(Q_1 = 812.4 \mathrm{~J}\)
02

Calculate the energy required for phase change of ice

Now, we will find the energy required to change the ice's phase from solid to liquid at 0°C using the formula Q = mL. We are given L = 6.01 kJ/mol. We need to convert this to J/g. Since the molar mass of water is 18 g/mol, L in J/g would be: \(L = \frac{6.01 \mathrm{~kJ/mol}}{18 \mathrm{~g/mol}} = 334 \mathrm{~J/g}\) Now use the formula with m = 20g, \(Q_2 = mL\) \(Q_2 = (20 \mathrm{~g})(334 \mathrm{~J/g})\) \(Q_2 = 6680 \mathrm{~J}\)
03

Calculate energy lost by water during the temperature change

The energy lost by water due to the ice will be equal to the energy gained by the ice, which is the sum of Q1 and Q2. \(Q_{total} = Q_1 + Q_2\) \(Q_{total} = 812.4 \mathrm{~J} + 6680 \mathrm{~J}\) \(Q_{total} = 7492.4 \mathrm{~J}\) Now using Q = mcΔT, we can find the change in temperature for the water (ΔT) when it loses Q_total amount of heat. We are given the mass of water, m = 500g (since its density is 1g/mL), and c = 4.184 J/g°C. \(\Delta T = \frac{Q_{total}}{mc}\) \(\Delta T = \frac{7492.4 \mathrm{~J}}{(500 \mathrm{~g})(4.184 \mathrm{~J/g°C})}\) \(\Delta T = 3.58°C\)
04

Calculate the final temperature of the mixture

Since water loses heat, its temperature will decrease by ΔT. The initial temperature of the water is 83°C. Therefore, the final temperature of the mixture will be: \(T_{final} = T_{initial} - \Delta T\) \(T_{final} = 83°C - 3.58°C\) \(T_{final} = 79.42°C\) The final temperature in the cup is approximately 79.42°C.

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Key Concepts

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

Enthalpy of Fusion
Understanding the enthalpy of fusion is crucial when studying the transition of substances from solid to liquid. This particular thermodynamic concept is defined as the amount of heat energy required to convert a solid into a liquid at constant pressure. Specifically, it applies to a phase change at the melting point of the substance.

In the provided exercise, the enthalpy of fusion is critical for calculating the energy needed to melt the ice. Given the enthalpy of fusion for water is 334 J/g, we can determine the heat energy necessary for the 20g ice cube to undergo phase change. This calculation provides insight into the energy dynamics during the melting process, which is an integral part of energy balance in thermochemical systems.
Specific Heat Capacity
Specific heat capacity is a property that indicates how much heat energy is required to raise the temperature of a unit mass of a substance by one degree Celsius. It's a fundamental concept in thermodynamics as it varies for different materials and states of matter.

The exercise demonstrates the use of specific heat capacity in calculating the energy needed to change the temperature of the ice and the water. Ice has a specific heat capacity of 2.03 J/g°C and water has a higher one at 4.184 J/g°C. Understanding these values enables us to predict how the ice and water will interact thermally when mixed and helps ascertain the final temperature of the system.
Heat Transfer Calculations
Heat transfer calculations are central to solving many thermochemistry problems. They involve quantifying the energy exchange that occurs due to temperature differences within a system or between systems. Commonly represented by the formula \(Q = mc\Delta T\), where \(Q\) stands for the heat transferred, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature.

In our exercise, these calculations are important for both understanding the warming of the ice and the cooling of the water. By applying the principles of conservation of energy, we can deduce that the heat lost by the water is equal to the heat gained by the ice, allowing us to calculate the final temperature of the water after equilibrium is reached.
Phase Change
Phase change refers to the process where a substance alters its physical state—such as transitioning from solid to liquid, or liquid to gas. This process occurs at specific temperatures and pressures for a given substance and often involves a significant amount of energy exchange, without changing the substance's temperature.

In the context of our problem, the phase change is happening as the ice melts. The energy needed for this phase change comes from the enthalpy of fusion of ice, which is already factored into our heat transfer calculations. Understanding the concept of phase change is vital, as it affects both the properties and behaviour of the materials involved, giving us a clearer picture of the thermal equilibrium in various processes.

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Most popular questions from this chapter

Which of the following processes are spontaneous and which are nonspontaneous: (a) the ripening of a banana, (b) dissolution of sugar in a cup of het coffee, (c) the reaction of nitrogen atoms to form \(\mathrm{N}_{2}\) molecules at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), (d) lightning. (c) formation of \(\mathrm{CH}_{4}\) and \(\mathrm{O}_{2}\) molecules from \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) at room temperature and \(1 \mathrm{~atm}\) of pressure?

(a) Using data in Appendix C, estimate the temperature at which the free- energy change for the transformation from \(I_{2}(s)\) to \(I_{2}(g)\) is zero. What assumptions must you make in arriving at this estimate? (b) Use a reference source, such as Web Elements (www.webelements.com), to find the experimental melting and boiling points of \(I_{2}\) (c) Which of the values in part (b) is closer to the value you obtained in part (a)? Can you explain why this is so?

Does the entropy of the system increase, decrease, or stay the same when (a) a solid melts, (b) a gas liquefies, (c) a solid sublimes?

Consider the vaporization of liquid water to steam at a pressure of \(1 \mathrm{~atm}\). (a) Is this process endothermic or exothermic? (b) In what temperature range is it a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) At what temperature are the two phases in equilibrium?

Indicate whether \(\Delta G\) increases, decreases, or does not change when the partial pressure of \(\mathrm{H}_{2}\) is increased in each of the following reactions: (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) \(2 \mathrm{HBr}(\mathrm{g}) \longrightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{Br}_{2}(g)\) (c) \(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})\)

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