/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 A \(237-\mathrm{g}\) piece of mo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(237-\mathrm{g}\) piece of molybdenum, initially at \(100.0^{\circ} \mathrm{C}\), is dropped into \(244 \mathrm{~g}\) water at \(10.0^{\circ} \mathrm{C}\). When the system comes to thermal equilibrium, the temperature is \(14.9^{\circ} \mathrm{C}\). Calculate the specific heat capacity of molybdenum.

Short Answer

Expert verified
The specific heat capacity of molybdenum is approximately \(0.249 \, \text{J/g}^\circ\text{C} \).

Step by step solution

01

Understanding the Problem

We need to find the specific heat capacity of molybdenum. We have a piece of molybdenum and some water coming to a thermal equilibrium at a certain temperature. The initial temperatures and masses are given for both substances.
02

User Energy Balance Equation

The heat lost by molybdenum must equal the heat gained by water when they come to thermal equilibrium. This can be expressed as: \( m_m c_m (T_{i,m} - T_f) = m_w c_w (T_f - T_{i,w}) \), where \( m_m \) and \( m_w \) are the masses, \( c_m \) is the specific heat of molybdenum, \( c_w \) is the specific heat of water, \( T_{i,m} \) and \( T_{i,w} \) are initial temperatures, and \( T_f \) is the final equilibrium temperature.
03

Substitute Known Values

The known values are: \( m_m = 237 \, \text{g} \), \( m_w = 244 \, \text{g} \), initial and final temperatures for molybdenum \( T_{i,m} = 100.0\, ^\circ\text{C} \) and \( T_f = 14.9\, ^\circ\text{C} \), and for water \( T_{i,w} = 10.0\, ^\circ\text{C} \) and \( T_f = 14.9\, ^\circ\text{C} \). The specific heat capacity of water \( c_w \) is \( 4.18 \, \text{J/g}^\circ\text{C} \). Insert these into the energy balance: \[ 237 \, c_m \, (100.0 - 14.9) = 244 \times 4.18 \times (14.9 - 10.0) \]
04

Solve for Specific Heat Capacity of Molybdenum

Calculate the left and right side of the equation. The right side calculates to \( 244 \times 4.18 \times 4.9 = 4981.68 \, \text{J} \). The left side becomes \( 237 \, c_m \, \times 85.1 \). Solving for \( c_m \) we get: \[ c_m = \frac{4981.68}{237 \times 85.1} \approx 0.249 \, \text{J/g}^\circ\text{C} \]
05

Conclusion and Verification

The specific heat capacity of molybdenum is approximately \(0.249 \, \text{J/g}^\circ\text{C} \). Verify the calculation matches with expected values, considering the typical range for metals, and ensure the energy conservation principle is met.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Thermal Equilibrium
When two substances interact thermally, they will exchange heat until they reach a uniform temperature known as thermal equilibrium. In simpler terms, thermal equilibrium occurs when no more heat flows between the substances. Each of them ends up at the same temperature.

In the exercise, a piece of molybdenum was mixed with water. Initially, the molybdenum was hotter than the water. As they interact, the molybdenum cools down while the water heats up. Eventually, both reach a common temperature of 14.9°C. This is their thermal equilibrium temperature.

Here are some key points about thermal equilibrium:
  • It involves energy transfer from a hotter object to a cooler one.
  • No net heat flow occurs at equilibrium.
  • It is an essential concept when analyzing heat transfer problems.
Understanding thermal equilibrium is crucial for problems involving specific heat capacity, as it helps determine the point at which heat transfer stops.
Energy Balance Equation
The energy balance equation is a central tool for solving problems involving heat transfer. It states that the heat lost by one substance should equal the heat gained by another. This happens when reaching thermal equilibrium, as in our exercise.

The exercise uses the energy balance equation: \( m_m c_m (T_{i,m} - T_f) = m_w c_w (T_f - T_{i,w}) \)

Here’s what this equation represents:
  • \( m_m \) and \( m_w \) are the masses of molybdenum and water, respectively.
  • \( c_m \) and \( c_w \) are their specific heat capacities.
  • \( T_{i,m} \) and \( T_{i,w} \) are their initial temperatures.
  • \( T_f \) is the final equilibrium temperature.
The equation shows how the heat lost by cooling molybdenum ( left side) equals the heat gained by warming water (right side). This mathematical expression is how we apply the principle of energy conservation—one loses what the other gains. By solving this equation, you can find unknowns, such as specific heat capacity, as demonstrated in the exercise.
Heat Transfer
Heat transfer is the movement of thermal energy from a warmer object to a cooler one. In the context of the problem, it is the process guiding the molybdenum and water to reach thermal equilibrium.

The three primary heat transfer methods are conduction, convection, and radiation. For this exercise, we’re most concerned with conduction, where heat transfers through direct contact. The molybdenum piece and water directly touching each other allow energy to pass through this method.

Important aspects of heat transfer include:
  • Heat always flows from the hotter object to the cooler one.
  • The rate of heat transfer depends on temperature difference, material properties, and contact surface.
  • Heat transfer will continue until both objects are at the same temperature.
Understanding these concepts is essential in solving thermal-related exercises. You can grasp how energy moves and how long it might take for objects to reach equilibrium.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What is required for heat transfer of energy from one sample of matter to another to occur?

A piece of iron ( \(400 . \mathrm{g}\) ) is heated in a flame and then plunged into a beaker containing \(1.00 \mathrm{~kg}\) water. The original temperature of the water was \(20.0^{\circ} \mathrm{C}\), but it is \(32.8^{\circ} \mathrm{C}\) after the iron bar is put in and thermal equilibrium is reached. Calculate the original temperature of the hot iron bar.

In some cities, taxicabs run on liquefied propane fuel instead of gasoline. This practice extends the lifetime of the vehicle and produces less pollution. Given that it costs about 2000 to modify the engine of a taxicab to run on propane and that the cost of gasoline and liquid propane are 3.50 per gallon and 2.50 per gallon, respectively, make reasonable assumptions and figure out how many miles a taxi would have to go so that the decreased fuel cost would balance the added cost of modifying the taxi's motor.

The sketch shows two identical beakers with different volumes of water at the same temperature. Is the thermal energy content of beaker 1 greater than, less than, or equal to that of beaker \(2 ?\) Explain your reasoning.

The specific heat capacity of copper is \(0.385 \mathrm{~J} \mathrm{~g}^{-1}{ }^{\circ} \mathrm{C}^{-1}\), whereas it is \(0.128 \mathrm{~J} \mathrm{~g}^{-1}{ }^{\circ} \mathrm{C}^{-1}\) for gold. Assume you place \(100 . \mathrm{g}\) of each metal, originally at \(25^{\circ} \mathrm{C},\) in a boiling water bath at \(100^{\circ} \mathrm{C}\). If energy is transferred to each metal at the same rate, determine which piece of metal will reach \(100^{\circ} \mathrm{C}\) first.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.