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Chapter 7: Problem 58

A \(110 .-\mathrm{g}\) sample of copper (specific heat capacity \(=\) the \(0.20 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\)) is heated to \(82.4^{\circ} \mathrm{C}\) and then placed in a container of water at \(22.3^{\circ} \mathrm{C} .\) The final temperature of the water and copper is \(24.9^{\circ} \mathrm{C} .\) What is the mass of the water in the container, assuming that all the heat lost by the copper is gained by the water?

Short Answer

Expert verified
The mass of the water in the container is approximately \(116.5 \,\text{g}\).

Step by step solution

01

Write down the formula for heat exchange

The formula for heat exchange is given by: \(q = mc螖T\) where: - \(q\) = heat exchanged - \(m\) = mass - \(c\) = specific heat capacity - \(螖T\) = change in temperature Since the heat lost by the copper is gained by the water, we can write the equation as: \(m_{Cu}c_{Cu}螖T_{Cu} = m_{H_2O}c_{H_2O}螖T_{H_2O}\), where the subscripts refer to copper and water.
02

Calculate the heat lost by the copper

We are given the following values for the copper: - \(m_{Cu} = 110 \,\text{g}\) - \(c_{Cu} = 0.20 \,\mathrm{J}/(\mathrm{g} \cdot ^{\circ}\mathrm{C})\) - \(T_{initial(Cu)} = 82.4 \,^{\circ}\mathrm{C}\) - \(T_{final(Cu)} = 24.9 \,^{\circ}\mathrm{C}\) The change in temperature for copper is: \(螖T_{Cu} = T_{final(Cu)} - T_{initial(Cu)} = 24.9 - 82.4 = -57.5^{\circ}\mathrm{C}\) Now, we can calculate the heat lost by the copper: \(q_{Cu} = m_{Cu} \cdot c_{Cu} \cdot 螖T_{Cu} = 110 \,\text{g} \cdot 0.20 \,\mathrm{J}/(\mathrm{g} \cdot ^{\circ}\mathrm{C}) \cdot (-57.5^{\circ}\mathrm{C}) = -1265 \,\text{J}\)
03

Calculate the heat gained by the water

Since the heat lost by the copper is gained by the water, we have: \(q_{H_2O} = -q_{Cu} = 1265 \,\text{J}\) We know the specific heat capacity of water: \(c_{H_2O} = 4.18 \,\mathrm{J}/(\mathrm{g} \cdot ^{\circ}\mathrm{C})\) and the initial and final temperature of the water: - \(T_{initial(H_2O)} = 22.3 \,^{\circ}\mathrm{C}\) - \(T_{final(H_2O)} = 24.9 \,^{\circ}\mathrm{C}\) The change in temperature for water is: \(螖T_{H_2O} = T_{final(H_2O)} - T_{initial(H_2O)} = 24.9 - 22.3 = 2.6^{\circ}\mathrm{C}\) Now we can calculate the mass of water in the container:
04

Calculate the mass of water

Using the equation \(q_{H_2O} =m_{H_2O} \cdot c_{H_2O} \cdot 螖T_{H_2O}\), we solve for the mass of water: \(m_{H_2O} = \dfrac{q_{H_2O}}{c_{H_2O} \cdot 螖T_{H_2O}} = \dfrac{1265\,\text{J}}{4.18 \,\mathrm{J}/(\mathrm{g} \cdot ^{\circ}\mathrm{C}) \cdot 2.6^{\circ}\mathrm{C}} \approx 116.5\, \mathrm{g}\) So, the mass of the water in the container is approximately \(116.5 \,\text{g}\).

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Key Concepts

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

Specific Heat Capacity
Specific heat capacity is a property that tells us how much heat energy is needed to raise the temperature of a given mass of a substance by one degree Celsius. It is denoted by the symbol 'c' and is measured in units of joules per gram per degree Celsius (\( \text{J/(g \textdegree C)} \) or \( \text{J/(kg \textdegree C)} \) if using kilograms).

The higher the specific heat capacity, the more energy is required to cause a change in temperature. For instance, water has a high specific heat capacity of 4.18 \text{J/(g \textdegree C)}, which means it can absorb a lot of heat before its temperature changes significantly. This property is crucial in heat exchange problems, as it establishes a relationship between heat added or lost and the corresponding temperature change. When we apply this concept to our problem, we use the known specific heat capacity of copper to determine the heat exchanged during its temperature drop.
Temperature Change in Heat Transfer
In the context of heat transfer, the temperature change (\( \text{螖T} \) represents the difference between the final temperature and the initial temperature of a substance. It is a crucial factor when calculating the heat absorbed or released by a substance during a thermal exchange. For positive values of \( \text{螖T} \) , the substance has absorbed heat, leading to an increase in temperature. Conversely, a negative \( \text{螖T} \) indicates heat loss and a temperature decrease.

In our textbook exercise, we see this concept in action as the copper cools down from \( 82.4\textdegree C \) to \( 24.9\textdegree C \) . This temperature change is used along with the specific heat capacity and mass to determine the amount of heat lost by the copper. Similarly, the water's temperature change from \( 22.3\textdegree C \) to \( 24.9\textdegree C \) allows us to calculate the heat it had gained.
Mass Calculation in Heat Exchange
Calculating the mass in a heat exchange problem often involves knowing the amount of heat transferred, the specific heat capacity, and the temperature change. The relationship between these variables is given by the formula \( q = mc\text{螖T} \) , where 'm' represents mass. By rearranging the formula, \( m = \frac{q}{c\text{螖T}} \) , we can solve for the unknown mass when the other values are known.

In our exercise, we look for the mass of the water that absorbs the heat lost by the copper. By using the calculated heat transfer (q) and the known specific heat capacity and temperature change for water, we arrive at the mass of the water. This step is critical for understanding heat exchange systems and energy balance in a variety of real-world applications, from everyday cooking to complex industrial processes.

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Most popular questions from this chapter

What is meant by the term lower in energy? Which is lower in energy, a mixture of hydrogen and oxygen gases or liquid water? How do you know? Which of the two is more stable? How do you know?

As a system increases in volume, it absorbs \(52.5 \mathrm{J}\) of energy in the form of heat from the surroundings. The piston is working against a pressure of \(0.500 \mathrm{atm}\). The final volume of the system is \(58.0 \mathrm{L}\). What was the initial volume of the system if the internal energy of the system decreased by \(102.5 \mathrm{J} ?\)

The bomb calorimeter in Exercise 102 is filled with 987 g water. The initial temperature of the calorimeter contents is \(23.32^{\circ} \mathrm{C} . \mathrm{A}\) \(1.056-\mathrm{g}\) sample of benzoic acid \(\left(\Delta E_{\text {comb }}=-26.42 \mathrm{kJ} / \mathrm{g}\right)\) is combusted in the calorimeter. What is the final temperature of the calorimeter contents?

Write reactions for which the enthalpy change will be a. \(\Delta H_{\mathrm{f}}^{\circ}\) for solid aluminum oxide. b. the standard enthalpy of combustion of liquid ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\). c. the standard enthalpy of neutralization of sodium hydroxide solution by hydrochloric acid. d. \(\Delta H_{\mathrm{f}}^{\circ}\) for gaseous vinyl chloride, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}(g)\). e. the enthalpy of combustion of liquid benzene, \(C_{6} \mathrm{H}_{6}(l)\). f. the enthalpy of solution of solid ammonium bromide.

Assuming gasoline is pure \(\mathrm{C}_{8} \mathrm{H}_{18}(l),\) predict the signs of \(q\) and \(w\) for the process of combusting gasoline into \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\).
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