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Chapter 19: Problem 15

(II) When a \(290-\mathrm{g}\) piece of iron at \(180^{\circ} \mathrm{C}\) is placed in a 95-g aluminum calorimeter cup containing \(250 \mathrm{~g}\) of glycerin at \(10^{\circ} \mathrm{C},\) the final temperature is observed to be \(38^{\circ} \mathrm{C}\) Estimate the specific heat of glycerin.

Short Answer

Expert verified
The specific heat capacity of glycerin is approximately \( 2.31 \text{ J/g}^\circ C \).

Step by step solution

01

Understand the Problem

We need to find the specific heat capacity of glycerin when a piece of iron is placed in an aluminum calorimeter containing glycerin. The initial and final temperatures of the substances are given.
02

Identify Given Variables

- Mass of iron, \( m_i = 290 \text{ g} \).- Initial temperature of iron, \( T_{i1} = 180^\circ C \).- Mass of aluminum calorimeter, \( m_a = 95 \text{ g} \).- Mass of glycerin, \( m_g = 250 \text{ g} \).- Initial temperature of glycerin and calorimeter, \( T_{g1} = T_{a1} = 10^\circ C \).- Final temperature, \( T_f = 38^\circ C \).- Specific heat of iron, \( c_i = 0.450 \text{ J/g}^\circ C \).- Specific heat of aluminum, \( c_a = 0.897 \text{ J/g}^\circ C \).
03

Write the Heat Transfer Equations

The heat lost by the iron will be equal to the heat gained by the glycerin and the calorimeter. Thus, \[ m_i c_i (T_{i1} - T_f) = m_g c_g (T_f - T_{g1}) + m_a c_a (T_f - T_{a1}) \]where \( c_g \) is the specific heat of glycerin we need to determine.
04

Plug in the Given Values

Substitute the known values into the equation:\[ 290 \times 0.450 \times (180 - 38) = 250 \times c_g \times (38 - 10) + 95 \times 0.897 \times (38 - 10) \]
05

Simplify and Solve for Specific Heat of Glycerin

Calculate terms step-by-step:- Left side: \( 290 \times 0.450 \times 142 = 18531 \text{ J} \).- Right side: Calculate the gain by aluminum: \( 95 \times 0.897 \times 28 = 2383.5 \text{ J} \).- Substitute and solve for \( c_g \): \[ 18531 = 250c_g imes 28 + 2383.5 \]\[ 250c_g imes 28 = 18531 - 2383.5 \]\[ c_g = \frac{16147.5}{250 \times 28} \approx 2.312 \text{ J/g}^\circ C \]
06

Conclusion

The specific heat capacity of glycerin is approximately \( 2.31 \text{ J/g}^\circ C \).

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Key Concepts

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

Calorimetry
Calorimetry is a branch of science that measures the heat of chemical reactions or physical changes. By using calorimetry, we can determine the amount of heat transferred to or from a substance. This process involves an instrument called a calorimeter.
A calorimeter is a simple device that often consists of an insulated container used to minimize the heat exchange with the environment. It typically contains a thermometer, a stirrer, and occasionally a heating device.
In the given exercise, an aluminum calorimeter cup is used. When the hot iron is placed inside, the calorimeter carefully captures the heat released by the iron and absorbed by the glycerin and the aluminum.
This observation allows us to apply the principle of heat transfer and analyze the data. By observing the change in temperature and knowing the masses and specific heat capacities of the components, we can calculate unknown quantities like the specific heat of glycerin.
  • Measure initial temperatures and masses.
  • Place substances into the calorimeter.
  • Allow them to reach thermal equilibrium.
  • Measure the final temperature.
Heat Transfer
Heat transfer is the process of heat energy moving from a hotter object to a cooler one until thermal equilibrium is reached. This movement can occur through conduction, convection, or radiation. In our exercise, conduction is primarily responsible as solid components are in direct contact with one another.

In the situation provided, the high-temperature iron piece transfers energy as heat to the cool glycerin and the aluminum calorimeter cup.
The heat lost by iron (\( Q = m_i c_i (T_{i1} - T_f) \) ), where it cools down, equals the heat absorbed by the glycerin and aluminum combined (\( Q = m_g c_g (T_f - T_{g1}) + m_a c_a (T_f - T_{a1}) \)).

The main aim is to demonstrate conservation of energy. All the heat energy lost by the iron must equal the total heat energy gained by the glycerin and the cup.
This balanced equation allows the calculation of specific quantities like specific heat, which can otherwise be difficult to determine directly, especially for liquids like glycerin.
  • Observe the initial and final energy states.
  • Conserve energy by equating heat lost and heat gained.
  • Identify conductive contact areas.
Thermal Equilibrium
Thermal equilibrium is the state where two substances reach the same temperature by exchanging heat. It indicates that there is no net heat transfer between the materials.

In our exercise, we observe thermal equilibrium when the final temperature of the system is reached. Initially, there is a temperature difference between the iron, the glycerin, and the aluminum calorimeter.
As the heat flows from the hotter iron to the cooler glycerin and aluminum, these substances gradually adjust until they stabilize at a uniform temperature.

At equilibrium, the different substances cease to transfer heat, indicating that they are all at the final temperature of 38掳C.
This principle is essential in calorimetry as it provides the data needed to solve equations of energy transfer.
  • Establish the condition when all components reach the same temperature.
  • Recognize the stopping of heat flow in the calorimeter.
  • Use equilibrium conditions to determine unknown properties.

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Most popular questions from this chapter

(II) A 3.65-mol sample of an ideal diatomic gas expands adiabatically from a volume of \(0.1210 \mathrm{~m}^{3}\) to \(0.750 \mathrm{~m}^{3}\). Initially the pressure was 1.00 atm. Determine: \((a)\) the initial and final temperatures; (b) the change in internal energy; (c) the heat lost by the gas; \((d)\) the work done on the gas. (Assume no molecular vibration.)

(II) The specific heat of mercury is \(138 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{C}^{\circ} .\) Determine the latent heat of fusion of mercury using the following calorimeter data: \(1.00 \mathrm{~kg}\) of solid \(\mathrm{Hg}\) at its melting point of \(-39.0^{\circ} \mathrm{C}\) is placed in a 0.620 -kg aluminum calorimeter with \(0.400 \mathrm{~kg}\) of water at \(12.80^{\circ} \mathrm{C}\); the resulting equilibrium temperature is \(5.06^{\circ} \mathrm{C}\).

(III) Consider a parcel of air moving to a different altitude \(y\) . in the Earth's atmosphere (Fig. \(33 ) .\) As the parcel changes altitude it acquires the pressure \(P\) of the surrounding air. From Eq. 20 we have $$\frac{d T}{d y}=-\rho g. $$where \(\rho\) is the parcel's altitude-dependent mass density. During this motion, the parcel's volume will change and, because air is a poor heat conductor, we assume this expansion or contraction will take place adiabatically. (a) Starting with Eq. 15, \(P V^{\gamma}=\) constant, show that for an ideal gas undergoing an adiabatic process, \(P^{1-\gamma} T^{\gamma}=\) constant. Then show that the parcel's pressure and temperature are related by $$(1-\gamma) \frac{d P}{d y}+\gamma \frac{P}{T} \frac{d T}{d y}=0$$ and thus $$(1-\gamma)(-\rho g)+\gamma \frac{P}{T} \frac{d T}{d y}=0$$ (b) Use the ideal gas law with the result from part \((a)\) to show that the change in the parcel's temperature with change in altitude is given by $$\frac{d T}{d y}=\frac{1-\gamma}{\gamma} \frac{m g}{k}$$ where \(m\) is the average mass of an air molecule and \(k\) is the Boltzmann constant. (c) Given that air is a diatomic gas with an average molecular mass of \(29,\) show that \(d T / d y=-9.8 \mathrm{C}^{\circ} / \mathrm{km} .\) This value is called the adiabatic lapse rate for dry air. (d) In California, the prevailing west- erly winds descend from one of the highest elevations (the 4000 -m Sierra Nevada mountains) to one of the lowest elevations (Death Valley, \(-100 \mathrm{m}\) ) in the continental United States. If a dry wind has a temperature of \(-5^{\circ} \mathrm{C}\) at the top of the Sierra Nevada, what is the wind's temperature after it has descended to Death Valley?

A house has well-insulated walls \(19.5 \mathrm{~cm}\) thick (assume conductivity of air) and area \(410 \mathrm{~m}^{2},\) a roof of wood \(5.5 \mathrm{~cm}\) thick and area \(280 \mathrm{~m}^{2}\), and uncovered windows \(0.65 \mathrm{~cm}\) thick and total area \(33 \mathrm{~m}^{2} .\) ( \(a\) ) Assuming that heat is lost only by conduction, calculate the rate at which heat must be supplied to this house to maintain its inside temperature at \(23^{\circ} \mathrm{C}\) if the outside temperature is \(-15^{\circ} \mathrm{C}\). \((b)\) If the house is initially at \(12^{\circ} \mathrm{C}\), estimate how much heat must be supplied to raise the temperature to \(23^{\circ} \mathrm{C}\) within 30 min. Assume that only the air needs to be heated and that its volume is \(750 \mathrm{~m}^{3}\). ( \(c\) ) If natural gas costs \(\$ 0.080\) per kilogram and its heat of combustion is \(5.4 \times 10^{7} \mathrm{~J} / \mathrm{kg},\) how much is the monthly cost to maintain the house as in part \((a)\) for \(24 \mathrm{~h}\) each day, assuming \(90 \%\) of the heat produced is used to heat the house? Take the specific heat of air to be \(0.24 \mathrm{kcal} / \mathrm{kg} \cdot \mathrm{C}^{\circ}\)

(II) A \(58-\mathrm{kg}\) ice-skater moving at \(7.5 \mathrm{~m} / \mathrm{s}\) glides to a stop. Assuming the ice is at \(0^{\circ} \mathrm{C}\) and that \(50 \%\) of the heat generated by friction is absorbed by the ice, how much ice melts?
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