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Chapter 7: Problem 4

A 75.0 g piece of \(\mathrm{Ag}\) metal is heated to \(80.0^{\circ} \mathrm{C}\) and dropped into \(50.0 \mathrm{g}\) of water at \(23.2^{\circ} \mathrm{C} .\) The final temperature of the \(\mathrm{Ag}-\mathrm{H}_{2} \mathrm{O}\) mixture is \(27.6^{\circ} \mathrm{C}\). What is the specific heat of silver?

Short Answer

Expert verified
The specific heat of silver is \(0.235 J/g掳C\).

Step by step solution

01

Calculate the Heat Gained by Water

First, let's calculate the heat gained by water using the formula \( q = mc\Delta T \). Here the mass \( m \) is \(50.0 g\), \( c \) (the specific heat capacity of water) is \( 4.18 J/g掳C \), and \(\Delta T \) (the change in temperature) is \( 27.6掳C - 23.2掳 C = 4.4掳 C \). Substituting these values into the equation, we find \( q = mc\Delta T = 50.0 g \times 4.18 J/g掳C \times 4.4 掳C = 921.92 J \). This is the heat gained by the water.
02

Calculate the Heat Lost by silver

Next, understand that the heat gained by the water is equal to the heat lost by silver. Thus, the heat lost by the silver, which we'll denote by \( q_{Ag} \), is also \(921.92 J\).
03

Calculate the Specific Heat of Silver

Finally, knowing the heat lost by the silver, you can calculate the specific heat of silver using the same formula \( q = mc\Delta T \). This time you know \( q_{Ag} = 921.92 J\), \( m = 75.0 g \), and \( \Delta T = 80掳C - 27.6掳C = 52.4掳C \). Plugging these values in the equation and solving for \( c \), you get: \( c = q / m\Delta T = 921.92 J / (75.0 g \times 52.4掳C ) = 0.235 J/g掳C \). The specific heat of silver is therefore \( 0.235 J/g掳C \).

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Key Concepts

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

Heat Transfer
Heat transfer is the process where energy in the form of heat moves from a hotter object to a cooler one. This happens naturally and can involve various methods such as conduction, convection, and radiation. In the context of our exercise with silver and water, conduction is the method of heat transfer. When the hot silver piece is dropped into cooler water, energy transfers from the silver to the water.
This continues until both reach the same temperature, known as thermal equilibrium. During this process:
  • The silver transfers heat because it initially has a higher temperature.
  • Water absorbs this heat, leading to an increase in its temperature.
  • The final temperature, once equilibrium is reached, is 27.6掳C.
This is an ongoing exchange governed by the principle of energy flow from hot to cold until balance is achieved.
Calorimetry
Calorimetry is the science of measuring the heat of chemical reactions or physical changes. It allows us to understand energy changes within a system. The concept is applied using devices called calorimeters, which are insulated to minimize energy exchange with surroundings.
In our exercise, we don't use an actual calorimeter, but we employ calorimetry principles:
  • The water and silver together can be considered a simple calorimeter system.
  • The heat lost by the silver (the hotter body) is equal to the heat gained by the water (the cooler body).
  • By accurately measuring temperature changes, we can calculate how much heat was transferred.
This practical application helps us determine properties like specific heat without a full calorimeter setup.
Energy Conservation
Energy conservation is a pivotal principle in physics, stating that energy cannot be created or destroyed, only transferred or converted from one form to another. In our exercise, this principle plays a crucial role.
When silver is placed in water, the total energy within the system remains constant.
  • Silver loses a quantifiable amount of heat energy.
  • Exactly this amount is what the water gains.
  • Through calculations, you find that 921.92 Joules is the heat transferred between these masses.
Thus, the exercise exhibits energy conservation plainly through the equal energy exchange between silver and water, highlighting that even in dynamic processes, the energy of a closed system remains unchanged.

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Most popular questions from this chapter

A calorimeter that measures an exothermic heat of reaction by the quantity of ice that can be melted is called an ice calorimeter. Now consider that \(0.100 \mathrm{L}\) of methane gas, \(\mathrm{CH}_{4}(\mathrm{g}),\) at \(25.0^{\circ} \mathrm{C}\) and \(744 \mathrm{mm} \mathrm{Hg}\) is burned at constant pressure in air. The heat liberated is captured and used to melt \(9.53 \mathrm{g}\) ice at \(0^{\circ} \mathrm{C}\left(\Delta H_{\text {fusion }} \text { of ice }=6.01 \mathrm{kJ} / \mathrm{mol}\right)\) (a) Write an equation for the complete combustion of \(\mathrm{CH}_{4},\) and show that combustion is incomplete in this case. (b) Assume that \(\mathrm{CO}(\mathrm{g})\) is produced in the incomplete combustion of \(\mathrm{CH}_{4}\), and represent the combustion as best you can through a single equation with small whole numbers as coefficients. \((\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) is another . product of the combustion.)

The following substances undergo complete combustion in a bomb calorimeter. The calorimeter assembly has a heat capacity of \(5.136 \mathrm{kJ} /^{\circ} \mathrm{C} .\) In each case, what is the final temperature if the initial water temperature is \(22.43^{\circ} \mathrm{C} ?\) \(\begin{array}{lllll}\text { (a) } 0.3268 & \text { g caffeine, } & \mathrm{C}_{8} \mathrm{H}_{10} \mathrm{O}_{2} \mathrm{N}_{4} & \text { (heat of }\end{array}\) combustion \(=-1014.2 \mathrm{kcal} / \mathrm{mol} \text { caffeine })\) (b) \(1.35 \mathrm{mL}\) of methyl ethyl ketone, \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}(1)\) \(d=0.805 \mathrm{g} / \mathrm{mL}\) (heat of combustion \(=-2444 \mathrm{kJ} / \mathrm{mol}\) methyl ethyl ketone).

What mass of ice can be melted with the same quantity of heat as required to raise the temperature of \(3.50 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}(1)\) by \(50.0^{\circ} \mathrm{C} ?\left[\Delta H_{\text {fusion }}^{\circ}=6.01 \mathrm{kJ} / \mathrm{mol}\right.\) \(\left.\mathrm{H}_{2} \mathrm{O}(\mathrm{s})\right]\)

A sample gives off 5228 cal when burned in a bomb calorimeter. The temperature of the calorimeter assembly increases by \(4.39^{\circ} \mathrm{C} .\) Calculate the heat capacity of the calorimeter, in kilojoules per degree Celsius.

Compressed air in aerosol cans is used to free electronic equipment of dust. Does the air do any work as it escapes from the can?
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