Skip to main content
Ch. 19 - Heat and the First Law of Thermodynamics
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
All textbooksGiancoli Douglas 5th editionCh. 19 - Heat and the First Law of ThermodynamicsProblem 28
Chapter 19, Problem 28

Determine the latent heat of fusion of mercury using the following calorimeter data: 1.00 kg of solid Hg at its melting point of −39.0°C is placed in a 0.620-kg aluminum calorimeter with 0.400 kg of water at 12.80°C; the resulting equilibrium temperature is 5.06°C.

Verified step by step guidance
1
Identify the principle of conservation of energy: The heat lost by the water and the aluminum calorimeter is equal to the heat gained by the mercury as it melts and warms to the equilibrium temperature.
Write the heat transfer equations for each component: (1) Heat lost by water: \( Q_{\text{water}} = m_{\text{water}} c_{\text{water}} (T_{\text{initial, water}} - T_{\text{final}}) \), (2) Heat lost by aluminum calorimeter: \( Q_{\text{calorimeter}} = m_{\text{calorimeter}} c_{\text{aluminum}} (T_{\text{initial, calorimeter}} - T_{\text{final}}) \), (3) Heat gained by mercury: \( Q_{\text{mercury}} = m_{\text{Hg}} L_f + m_{\text{Hg}} c_{\text{Hg}} (T_{\text{final}} - T_{\text{melting, Hg}}) \).
Substitute the known values into the equations: \( m_{\text{water}} = 0.400 \ \text{kg}, \ c_{\text{water}} = 4186 \ \text{J/kg·°C}, \ T_{\text{initial, water}} = 12.80 \ \text{°C}, \ T_{\text{final}} = 5.06 \ \text{°C} \); \( m_{\text{calorimeter}} = 0.620 \ \text{kg}, \ c_{\text{aluminum}} = 900 \ \text{J/kg·°C}, \ T_{\text{initial, calorimeter}} = 12.80 \ \text{°C}, \ T_{\text{final}} = 5.06 \ \text{°C} \); \( m_{\text{Hg}} = 1.00 \ \text{kg}, \ T_{\text{melting, Hg}} = -39.0 \ \text{°C}, \ c_{\text{Hg}} = 140 \ \text{J/kg·°C} \).
Set up the energy balance equation: \( Q_{\text{water}} + Q_{\text{calorimeter}} = Q_{\text{mercury}} \). Expand this equation using the heat transfer expressions: \( m_{\text{water}} c_{\text{water}} (T_{\text{initial, water}} - T_{\text{final}}) + m_{\text{calorimeter}} c_{\text{aluminum}} (T_{\text{initial, calorimeter}} - T_{\text{final}}) = m_{\text{Hg}} L_f + m_{\text{Hg}} c_{\text{Hg}} (T_{\text{final}} - T_{\text{melting, Hg}}) \).
Solve for the latent heat of fusion \( L_f \) of mercury: Rearrange the equation to isolate \( L_f \): \( L_f = \frac{[m_{\text{water}} c_{\text{water}} (T_{\text{initial, water}} - T_{\text{final}}) + m_{\text{calorimeter}} c_{\text{aluminum}} (T_{\text{initial, calorimeter}} - T_{\text{final}}) - m_{\text{Hg}} c_{\text{Hg}} (T_{\text{final}} - T_{\text{melting, Hg}})]}{m_{\text{Hg}}} \). Substitute the known values and simplify to find \( L_f \).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
8m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Latent Heat

Latent heat is the amount of heat energy required to change a substance from one phase to another without changing its temperature. In the case of fusion, it refers to the energy needed to convert a solid into a liquid at its melting point. This concept is crucial for understanding how energy is transferred during phase changes, such as when solid mercury melts.
Recommended video:
Guided course
10:40
Latent Heat & Phase Changes

Calorimetry

Calorimetry is the science of measuring the heat of chemical reactions or physical changes. In this context, it involves calculating the heat exchanged between the solid mercury, the aluminum calorimeter, and the water to determine the latent heat of fusion. Understanding calorimetry principles is essential for analyzing thermal equilibrium and energy conservation in the system.
Recommended video:
Guided course
06:53
Solving Calorimetry Problems

Thermal Equilibrium

Thermal equilibrium occurs when two or more bodies in contact with each other reach the same temperature, resulting in no net heat flow between them. In this problem, the final equilibrium temperature of 5.06°C indicates that heat has been exchanged among the solid mercury, water, and the calorimeter until they all reached this common temperature. This concept is fundamental for applying the principles of energy conservation in calorimetry.
Recommended video:
Guided course
05:21
Volume Thermal Expansion
Related Practice
Textbook Question

An iron boiler of mass 180 kg contains 710 kg of water at 18°C. A heater supplies energy at the rate of 58,000 kJ/h. How long does it take for the water to all have changed to steam?

Textbook Question

At a crime scene, the forensic investigator notes that the 6.2-g lead bullet that was stopped in a doorframe apparently melted completely on impact. Assuming the bullet was shot at room temperature (20°C), what does the investigator calculate as the minimum muzzle velocity of the gun?

1
views
Textbook Question

A 1.0-L volume of air initially at 3.5 atm of (gauge)pressure is allowed to expand isothermally until the (gauge) pressure is 1.0 atm. It is then compressed at constant pressure to its initial volume, and lastly is brought back to its original pressure by heating at constant volume. How much work does the 1.0 L of air do in this process?

Textbook Question

High-altitude mountain climbers do not eat snow, but always melt it first with a stove. To see why, calculate the energy absorbed from your body if you melt 1.0 kg of -15°C snow using a stove and drink the resulting 1.0 kg of water at 2°C, which your body has to warm to 37°C.

1
views
Textbook Question

What mass of steam at 100°C must be added to 1.00 kg of ice at 0°C to yield liquid water at 30°C?

1
views