Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A piece of titanium metal with a mass of \(20.8 \mathrm{g}\) is heated in boiling water to \(99.5^{\circ} \mathrm{C}\) and then dropped into a coffee-cup calorimeter containing \(75.0 \mathrm{g}\) of water at \(21.7^{\circ} \mathrm{C} .\) When thermal equilibrium is reached, the final temperature is \(24.3^{\circ} \mathrm{C} .\) Calculate the specific heat capacity of titanium.

Short Answer

Expert verified
The specific heat capacity of titanium is approximately 0.52 J/g°C.

Step by step solution

01

Identify known values

First, let's list what we know from the problem: - Mass of titanium ( mt ) = 20.8 g - Initial temperature of titanium ( T_{i,t} ) = 99.5 °C - Mass of water ( mw ) = 75.0 g - Initial temperature of water ( T_{i,w} ) = 21.7 °C - Final temperature of both systems ( T_{f} ) = 24.3 °C - Specific heat capacity of water ( cw ) = 4.18 J/g°C.
02

Write the energy balance equation

Since energy is conserved, the heat lost by titanium equals the heat gained by water. Thus, we can write:\[m_t c_t (T_{i,t} - T_f) = m_w c_w (T_f - T_{i,w})\]where mt is the mass of titanium, c_t is the specific heat capacity of titanium, T_{i,t} is the initial temperature of titanium, T_f is the final temperature, T_{i,w} is the initial temperature of water, m_w is the mass of water, and c_w is the specific heat capacity of water.
03

Substitute known values and solve for specific heat capacity

Insert the known values into the equation:\[20.8 \, \text{g} \cdot c_t \cdot (99.5 - 24.3) = 75.0 \, \text{g} \cdot 4.18 \, \frac{\text{J}}{\text{g} \, ^\circ \text{C}} \cdot (24.3 - 21.7)\]Calculate both sides,- Left side: \[20.8 \, \text{g} \cdot c_t \cdot 75.2\]- Right side: \[75.0 \, \text{g} \cdot 4.18 \, \frac{\text{J}}{\text{g} \, ^\circ \text{C}} \cdot 2.6^\circ \text{C} = 815.4 \, \text{J}\]Set the equations equal and solve for c_t :\[20.8 \, \text{g} \cdot c_t \cdot 75.2 = 815.4\] \[c_t = \frac{815.4}{20.8 \cdot 75.2}\]Calculate c_t to find the specific heat capacity of titanium.
04

Calculate and interpret result

Perform the final calculation:\[c_t = \frac{815.4}{1565.76}\approx 0.52 \, \text{J/g°C}\]This means that the specific heat capacity of titanium is approximately 0.52 J/g°C.

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 Vaia!

Key Concepts

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

Thermal Equilibrium
Thermal equilibrium is a vital concept in understanding how heat exchange works between different objects. When two objects at different temperatures come into contact, they will exchange heat energy until they reach the same temperature. This shared temperature is known as the thermal equilibrium. In the given exercise, titanium and water equilibrate to a final temperature of \(24.3^{\circ} \text{C}\).
The energy lost by the hotter titanium is equal to the energy gained by the cooler water. This is because no heat is lost to the surroundings in an ideal calorimetry scenario. Once both substances reach the same temperature, heat transfer ceases. Thermal equilibrium helps us predict the final temperature and set up equations needed for calculating specific heat values.
Calorimetry
Calorimetry is the science of measuring the change in heat associated with chemical reactions or physical changes. In this problem, a coffee-cup calorimeter is used to measure heat exchange between titanium and water. Calorimeters help us by isolating the process to ensure accurate heat transfer measurement without external heat loss or gain.
In the exercise, we determine the specific heat of titanium by monitoring how it exchanges heat with water, whose specific heat is known. Calorimetry allows us to apply equations that represent the amount of heat transferred, connecting the masses, specific heats, and temperature changes of the substances involved.
  • Mass and specific heat: These are crucial for determining how much heat an object can hold.
  • Temperature changes: They indicate the amount of heat transfer involved in reaching thermal equilibrium.
Through careful measurement and proper setup, calorimetry provides a precise way to calculate specific heat capacities, such as that of titanium in this exercise.
Energy Conservation
Energy conservation is a fundamental concept in physics, stating that energy cannot be created or destroyed, only transformed. When dealing with calorimetry, this principle is crucial. The heat lost by one material must be equal to the heat gained by another when they exchange energy.
In the problem, energy conservation is mathematically expressed as the equivalence between the heat loss of titanium and the heat gain of water. Setting these two quantities equal allows us to solve for the unknown specific heat capacity of titanium.
  • Identify known variables such as mass, specific heat, and temperatures.
  • Use the energy balance equation to equate the heat transfer.
  • Solve for the specific heat of the unknown material.
This method underscores the power of energy conservation in predicting and calculating the behavior of materials during temperature change.

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

The value of \(\Delta U\) for the decomposition of \(7.647 \mathrm{g}\) of ammonium nitrate can be measured in a bomb calorimeter. The reaction that occurs is $$ \mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{s}) \rightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$The temperature of the calorimeter, which contains \(415 \mathrm{g}\) of water, increases from \(18.90^{\circ} \mathrm{C}\) to \(20.72^{\circ} \mathrm{C} .\) The heat capacity of the bomb is \(155 \mathrm{J} / \mathrm{K}\). What is the value of \(\Delta U\) for this reaction, in \(\mathrm{kJ} / \mathrm{mol}\) ? (IMAGE CAN'T COPY)

Suppose you burned \(1.500 \mathrm{g}\) of benzoic acid, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H},\) in a constant volume calorimeter and found that the temperature increased from \(22.50^{\circ} \mathrm{C}\) to \(31.69^{\circ} \mathrm{C} .\) The calorimeter contained \(775 \mathrm{g}\) of water, and the bomb had a heat capacity of \(893 \mathrm{J} / \mathrm{K}\). Calculate \(\Delta U\) per mole of benzoic acid. (IMAGE CAN'T COPY)

Which of the following are state functions? (a) the volume of a balloon (b) the time it takes to drive from your home to your college or university (c) the temperature of the water in a coffee cup (d) the potential energy of a ball held in your hand

The specific heat capacity of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) is \(1.74 \mathrm{J} / \mathrm{g} \cdot \mathrm{K} .\) What is its molar heat capacity (in \(\mathrm{J} / \mathrm{mol} \cdot \mathrm{K}) ?\)

How much energy is required to vaporize \(125 \mathrm{g}\) of benzene, \(\mathrm{C}_{6} \mathrm{H}_{6},\) at its boiling point, \(80.1^{\circ} \mathrm{C} ?\) (The heat of vaporization of benzene is \(30.8 \mathrm{kJ} / \mathrm{mol} .\) )

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.

Sign-up for free