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Chapter 6: Problem 58

A \(110 .-\mathrm{g}\) sample of copper (specific heat capacity \(=0.20 \mathrm{~J} /{ }^{\circ} \mathrm{C}\). \(\mathrm{g}\) ) 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 \(72.5\ \mathrm{g}\).

Step by step solution

01

Write down the known variables

From the exercise, we know the following: - The mass of copper (m_copper) = 110 g - The specific heat capacity of copper (c_copper) = 0.20 J/g掳C - The initial temperature of copper (T_initial_copper) = 82.4掳C - The initial temperature of water (T_initial_water) = 22.3掳C - The final temperature of both copper and water (T_final_both) = 24.9掳C
02

Write the equation for heat transfer

We know that when we put the hot copper inside the container, the copper will lose heat and the water will gain heat until both reach the same final temperature. So, we can use the following equation for heat transfer when all the heat lost by the copper is gained by the water: \(Q_{lost\_copper} = Q_{gained\_water}\)
03

Find the heat lost by copper

We can find the heat lost by the copper using the equation: \(Q_{lost\_copper} = m_{copper} \times c_{copper} \times \Delta T_{copper}\) Where: - m_copper = mass of copper - c_copper = specific heat capacity of copper - 鈭員_copper = change in temperature of copper = |T_final_both - T_initial_copper| Now, substituting the values, we get: \(Q_{lost\_copper} = (110) \times (0.20) \times (|24.9 - 82.4|)\) Calculating the heat lost by copper: \(Q_{lost\_copper} = 110 \times 0.20 \times 57.5 = 1265\ \mathrm{J}\)
04

Find the mass of water using heat gained by water

Now, we will find the mass of water using the heat gained by water: \(Q_{gained\_water} = m_{water} \times c_{water} \times \Delta T_{water}\) Where: - m_water = mass of water - c_water = specific heat capacity of water = 4.18 J/g掳C (given) - 鈭員_water = change in temperature of water = |T_final_both - T_initial_water| Since we know that \(Q_{lost\_copper} = Q_{gained\_water}\), we can rewrite the equation as: \(m_{water} = \frac{Q_{gained\_water}}{c_{water} \times \Delta T_{water}}\) Substituting the values, we get: \(m_{water} = \frac{1265}{(4.18) \times (|24.9 - 22.3|)}\) Calculating the mass of water: \(m_{water} = \frac{1265}{4.18 \times 2.6} = 72.5\ \mathrm{g}\) So, the mass of the water in the container is 72.5 g.

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Key Concepts

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

Specific Heat Capacity
Understanding the concept of specific heat capacity is crucial in solving problems involving heat transfer. Specific heat capacity is a measure of the amount of thermal energy required to change the temperature of a unit mass of a substance by 1 degree Celsius. It helps us understand how different materials react to heat.

For example, copper has a relatively low specific heat capacity of 0.20 J/g掳C. This means it doesn't require much energy to increase its temperature. On the other hand, water has a higher specific heat capacity of 4.18 J/g掳C, indicating that it can absorb more heat without a significant change in temperature. This difference explains why water is effective in moderating temperature changes in various applications.

When applying this concept to exercises, remember:
  • Low specific heat capacity means quicker temperature changes with heat addition or removal.
  • High specific heat capacity means slower temperature changes with heat addition or removal.
  • It's essential to know the specific heat capacity values to calculate heat transfer effectively.
Temperature Change
Temperature change is at the core of analyzing heat transfer between substances. It describes the difference between the initial and final temperatures, denoted as \(\Delta T\). In the given problem, we see temperature changes in both copper and water.

The copper starts at 82.4掳C and cools down to 24.9掳C, resulting in a temperature change of \(|24.9 - 82.4| = 57.5掳C\). For water, the change is from 22.3掳C to 24.9掳C, equating to \(|24.9 - 22.3| = 2.6掳C\). Understanding how to calculate these changes is crucial for solving heat transfer problems, as they dictate the amount of heat gained or lost by each substance.

When calculating temperature changes, remember:
  • Always subtract the final temperature from the initial temperature to get the magnitude of change.
  • Use absolute values as temperature changes cannot be negative in this context.
  • Temperature change provides a direct input into calculating the energy transferred during heating/cooling processes.
Thermal Equilibrium
Thermal equilibrium is a state when two or more linked systems reach a common temperature and no net energy transfer occurs between them. This concept is essential in understanding how heat flows from hotter objects to cooler ones until equilibrium is achieved.

In the exercise, copper and water eventually reach a final temperature of 24.9掳C. They achieve thermal equilibrium as the copper loses heat and the water gains it until both are at the same temperature. This equilibrium state is vital for solving such problems because it implies that the heat lost by one component is equal to the heat gained by the other.

Important points about thermal equilibrium include:
  • The principle that energy flows from hot to cold until both objects are equally warm.
  • The idea that net energy change (total heat transfer) is zero at equilibrium.
  • Ensures conservation of energy during heat exchange, allowing accurate calculation of unknown values, like mass, in heat transfer problems.

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Most popular questions from this chapter

Given the following data $$ \begin{aligned} \mathrm{P}_{4}(s)+6 \mathrm{Cl}_{2}(g) & \longrightarrow 4 \mathrm{PCl}_{3}(g) & & \Delta H=-1225.6 \mathrm{~kJ} \\ \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) & \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) & & \Delta H=-2967.3 \mathrm{~kJ} \\ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) & \longrightarrow \mathrm{PCl}_{5}(g) & & \Delta H=-84.2 \mathrm{~kJ} \\ \mathrm{PCl}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) & \mathrm{Cl}_{3} \mathrm{PO}(g) & & \Delta H=-285.7 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the reaction $$ \mathrm{P}_{4} \mathrm{O}_{10}(s)+6 \mathrm{PCl}_{5}(g) \longrightarrow 10 \mathrm{Cl}_{3} \mathrm{PO}(g) $$

Consider the substances in Table 6.1. Which substance requires the largest amount of energy to raise the temperature of \(25.0 \mathrm{~g}\) of the substance from \(15.0^{\circ} \mathrm{C}\) to \(37.0^{\circ} \mathrm{C}\) ? Calculate the energy. Which substance in Table \(6.1\) has the largest temperature change when \(550 . \mathrm{g}\) of the substance absorbs \(10.7 \mathrm{~kJ}\) of energy? Calculate the temperature change.

The bomb calorimeter in Exercise 108 is filled with \(987 \mathrm{~g}\) water. The initial temperature of the calorimeter contents is \(23.32^{\circ} \mathrm{C}\). 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?

Consider the following reaction: $$ 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=-572 \mathrm{~kJ} $$ a. How much heat is evolved for the production of \(1.00 \mathrm{~mol}\) \(\mathrm{H}_{2} \mathrm{O}(l) ?\) b. How much heat is evolved when \(4.03 \mathrm{~g}\) hydrogen is reacted with excess oxygen? c. How much heat is evolved when \(186 \mathrm{~g}\) oxygen is reacted with excess hydrogen? d. The total volume of hydrogen gas needed to fill the Hindenburg was \(2.0 \times 10^{8} \mathrm{~L}\) at \(1.0 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\). How much heat was evolved when the Hindenburg exploded, assuming all of the hydrogen reacted?

How is average bond strength related to relative potential energies of the reactants and the products?
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