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Assuming no heat loss by the system, what will be the final temperature when \(50.0 \mathrm{~g}\) of water at \(10.0^{\circ} \mathrm{C}\) are mixed with \(10.0 \mathrm{~g}\) of water at \(50.0^{\circ} \mathrm{C}\) ?

Short Answer

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50.0°°ä

Step by step solution

01

Understand the Concept

To solve this problem, use the principle of conservation of energy, which implies that the total heat lost by the warmer water will equal the total heat gained by the cooler water. The equation for heat exchange is: \(Q = mc\Delta T\), where Q is the heat energy, m is the mass, c is the specific heat capacity (for water it's \(4.18 \mathrm{\frac{J}{g\cdot ^\circ C}}\)), and \(\Delta T\) is the change in temperature.
02

Set Up the Equation

Let \(T_f\) be the final temperature of the mixture. For the warmer water losing heat, \(Q_{lost} = m_{hot}c\Delta T_{hot}\) and for the cooler water gaining heat, \(Q_{gained} = m_{cold}c\Delta T_{cold}\). Since no heat is lost to the surroundings, the heat lost by the hot water is equal to the heat gained by the cold water: \(m_{hot}c(T_{hot} - T_f) = m_{cold}c(T_f - T_{cold})\).
03

Insert the Known Values

Plug in the given masses and temperatures to solve for \(T_f\): \(10.0 \mathrm{~g} \cdot 4.18 \mathrm{\frac{J}{g\cdot ^\circ C}} \cdot (50.0^\circ \mathrm{C} - T_f) = 50.0 \mathrm{~g} \cdot 4.18 \mathrm{\frac{J}{g\cdot ^\circ C}} \cdot (T_f - 10.0^\circ \mathrm{C})\).
04

Simplify the Equation

The specific heat capacity cancels out, simplifying the equation to \(10.0 \cdot (50.0 - T_f) = 50.0 \cdot (T_f - 10.0)\).
05

Expand and Solve for \(T_f\)

After expansion, the equation becomes \(500.0 - 10.0 T_f = 50.0 T_f - 500.0\). Rearrange to solve for \(T_f\) by adding \(10.0 T_f\) to both sides and adding \(500.0\) to both sides, to get \(20.0 T_f = 1000.0\). Divide both sides by \(20.0\) to find the final temperature, \(T_f\).
06

Calculate the Final Temperature

By dividing both sides by \(20.0\), we get \(T_f = \frac{1000.0}{20.0} = 50.0^\circ \mathrm{C}\).

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Key Concepts

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

Conservation of Energy
The principle of conservation of energy is a fundamental concept in physics and chemistry. It states that energy cannot be created or destroyed in an isolated system. When it comes to heat exchange in chemistry, the conservation of energy means that the total amount of energy in the system remains constant, even as energy may change from one form to another.

In the context of mixing two masses of water at different temperatures, as seen in the problem statement, the warmer water loses heat energy while the cooler water gains heat energy. The loss and gain of energy must be equal if no heat is lost to the surroundings, resulting in a new common temperature where the energies are balanced. This conservation principle allows us to set up equations that facilitate finding the final temperature after mixing substances, ensuring the total energy before and after mixing remains the same.
Specific Heat Capacity
Specific heat capacity is a property that describes how much heat energy is needed to raise the temperature of a substance by one degree Celsius.

Mathematically, it is expressed as: \[\begin{equation} c = \frac{Q}{m\triangle T} \<\br> \<\br> \<\br> \<\br> \<\br> \<\br> \<\br>\end{equation}\] where
  • \( Q \) is the amount of heat energy in joules (J),
  • \( m \) is the mass of the substance in grams (g), and
  • \( \triangle T \) is the change in temperature in degrees Celsius (\textdegree{}C).
The specific heat capacity of water, used in the exercise, is relatively high, which means it requires a lot of energy to change its temperature. This property is crucial when calculating the final temperature in heat exchange problems because different substances will respond differently to the same amount of heat energy.
Temperature Equilibration
Temperature equilibration is the process by which two substances in contact with each other exchange heat until they reach the same temperature. This concept is related to the zeroth law of thermodynamics, which states that if two systems are each in thermal equilibrium with a third system, they are in thermal equilibrium with each half.

During temperature equilibration, heat flows from the hotter substance to the cooler one. In our exercise, the heat transferred from the warmer water to the cooler water results in a decrease and increase in temperature respectively until both reach a final equilibrium temperature. The final temperature can be computed assuming no heat is lost to the environment, which is an ideal scenario often used in chemistry problems to simplify calculations.
Chemical Thermodynamics
Chemical thermodynamics is the branch of chemistry that deals with the study of energy changes, particularly the conversion of heat energy into work and vice versa, during chemical reactions and physical changes. It encompasses the laws of thermodynamics, which guide the behavior of systems in terms of energy transfer.

In the scope of our discussion, when we talk about heat exchange between warm and cold water, we are dealing with a fundamental thermodynamic process not involving chemical reactions, but the concepts can be extended to such. These processes are guided by the same laws. The first law of thermodynamics, equivalent to the conservation of energy principle, ensures that in a closed system the energy change in the system must be equal and opposite to the energy change in the surroundings. By mastering the basics of heat exchange and thermodynamics, students can apply the same principles to understand more complex chemical systems.

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Most popular questions from this chapter

State whether each of the following represents a chemical change or a physical change: (a) A few grams of sucrose (table sugar) are placed in a small beaker of deionized water; the sugar crystals "disappear," and the liquid in the beaker remains clear and colorless. (b) A copper statue, over time, turns green. (c) When a teaspoon of baking soda (sodium bicarbonate) is placed into a few ounces of vinegar (acetic acid), volumes of bubbles (effervescence) are produced. (d) When a few grams of a blue crystalline solid are placed into a beaker of deionized water, the crystals "disappear" and the liquid becomes clear and blue in color. (e) In the lab, a student mixes \(2 \mathrm{~mL}\) of sodium hydroxide with \(2 \mathrm{~mL}\) of hydrochloric acid in a test tube. He notices that the test tube is very warm to the touch. (f) A woman visits a hairdresser and has her hair colored a darker shade of brown. After several weeks the hair, even though washed several times, does not change back to the original color.

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