/*! 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 60 A glass vial containing a 16.0 -... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 60

A glass vial containing a 16.0 -g sample of an enzyme is cooled in an ice bath. The bath contains water and 0.120 \(\mathrm{kg}\) of ice. The sample has specific heat 2250 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) ; the glass vial has mass 6.00 \(\mathrm{g}\) and specific heat 2800 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) . How much ice melts in cooling the enzyme sample from room temperature \(\left(19.5^{\circ} \mathrm{C}\right)\) to the temperature of the ice bath?

Short Answer

Expert verified
3.08 grams of ice melts.

Step by step solution

01

Understand the Problem

We need to calculate how much ice melts when a 16.0-g enzyme sample, initially at 19.5掳C, is cooled in an ice bath. We will also consider a 6.00-g glass vial. The specific heat capacities of the enzyme and the glass vial are provided. We need to find the heat transfer from the enzyme and vial to the ice and determine how much ice melts.
02

Calculate Heat Lost by Enzyme

The formula for heat lost is given by: \[ q = m \cdot c \cdot \Delta T \]where \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the temperature change. For the enzyme:\[ m = 0.016 \, \text{kg}, \quad c = 2250 \, \text{J/kg} \cdot \text{K}, \quad \Delta T = 19.5 - 0 = 19.5 \text{K} \]Calculate:\[ q_{\text{enzyme}} = 0.016 \cdot 2250 \cdot 19.5 = 702 \, \text{J} \]
03

Calculate Heat Lost by Glass Vial

Similarly, calculate the heat lost by the glass vial using the same formula:\[ m = 0.006 \, \text{kg}, \quad c = 2800 \, \text{J/kg} \cdot \text{K}, \quad \Delta T = 19.5 \]Calculate:\[ q_{\text{vial}} = 0.006 \cdot 2800 \cdot 19.5 = 327.6 \, \text{J} \]
04

Total Heat Loss

Add the heat lost by the enzyme and the glass vial to get the total heat loss:\[ q_{\text{total}} = q_{\text{enzyme}} + q_{\text{vial}} = 702 + 327.6 = 1029.6 \, \text{J} \]
05

Calculate Mass of Melted Ice

The heat required to melt ice is given by:\[ q = m \cdot L_f \]where \( L_f = 334,000 \, \text{J/kg} \) is the latent heat of fusion of ice. Rearrange the formula to find the mass:\[ m = \frac{q}{L_f} = \frac{1029.6}{334,000} \approx 0.00308 \, \text{kg} \]Convert to grams:\[ 0.00308 \, \text{kg} = 3.08 \, \text{g} \]

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 91影视!

Key Concepts

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

Heat Transfer
Heat transfer is a fundamental concept in thermodynamics. It involves the movement of thermal energy from one object or substance to another. This transfer occurs due to a temperature difference between the systems. There are three main modes of heat transfer: conduction, convection, and radiation.
  • Conduction is the transfer of heat through a material without any flow of the material itself, such as heating a metal rod at one end.
  • Convection involves the movement of heat by the physical movement of fluid from one location to another, like the circulation within a pot of boiling water.
  • Radiation is the transfer of heat through electromagnetic waves, such as the warmth of sunlight.
In our exercise, heat is transferred from the warmed enzyme sample to the cold ice bath until thermal equilibrium is achieved. This results in the melting of some ice, as the transferred heat causes a phase change.
Specific Heat Capacity
Specific heat capacity is a property that indicates the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius. It is a critical factor when calculating heat transfer during temperature changes.
  • The formula to calculate the heat transfer when there is a change in temperature is: \[ q = m \cdot c \cdot \Delta T \]
    where:
    • \( q \) is the heat energy (in joules),
    • \( m \) is the mass (in kilograms),
    • \( c \) is the specific heat capacity (in joules per kilogram per degree Celsius),
    • \( \Delta T \) is the temperature change (in degrees Celsius).
In the provided solution, specific heat capacities for the enzyme and the glass vial are used to calculate how much heat each loses as they cool down to the temperature of the ice bath. The values \( 2250 \, \text{J/kg} \cdot \text{K} \) for the enzyme and \( 2800 \, \text{J/kg} \cdot \text{K} \) for the vial show how different materials require differing amounts of energy to change their temperatures.
Latent Heat
Latent heat refers to the amount of heat that is required to change the state of a substance without changing its temperature. There are different types of latent heat, such as latent heat of fusion, vaporization, etc.
  • For melting ice, we use the latent heat of fusion, which is the heat needed to convert ice at 0掳C to water at 0掳C without temperature change.
  • The formula for latent heat is: \[ q = m \cdot L_f \]
    where:
    • \( q \) is the heat energy required for the phase change (in joules),
    • \( m \) is the mass (in kilograms),
    • \( L_f \) is the latent heat of fusion (in joules per kilogram).
In this exercise, when calculating the amount of ice melted, the latent heat of fusion value used is \( 334,000 \, \text{J/kg} \). It determines how much energy is needed to convert a given mass of ice to water.
Phase Change
Phase change refers to the transformation of a substance from one state of matter to another, such as solid to liquid, liquid to gas, etc. These changes occur at specific temperatures and involve energy transfer without changing the temperature of the substance during the process.
  • Common phase changes include melting, freezing, condensation, boiling, and evaporation.
  • For example, when ice melts, it undergoes a transition from a solid to a liquid state.
  • This change requires energy input in the form of heat, known as latent heat.
In our context, the glass vial and enzyme as they cool, transfer heat to the ice bath. This energy contributes to the phase change from solid (ice) to liquid (water), shown by the calculated melting of 3.08 grams of ice. Understanding phase change is crucial for assessing how much heat energy in systems like this is used in state transformations versus 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

A vessel whose walls are thermally insulated contains 2.40 \(\mathrm{kg}\) of water and 0.450 \(\mathrm{kg}\) of ice, all at a temperature of \(0.0^{\circ} \mathrm{C}\) The outlet of a tube leading from a boiler in which water is boiling at atmospheric pressure is inserted into the water. How many grams of steam must condense inside the vessel (also at atmospheric pressure) to raise the temperature of the system to \(28.0^{\circ} \mathrm{C}\) ? You can ignore the heat transferred to the container.

(a) If an area measured on the surface of a solid body is \(A_{0}\) at some initial temperature and then changes by \(\Delta A\) when the temperature changes by \(\Delta T,\) show that $$\Delta A=(2 \alpha) A_{0} \Delta T$$ where \(\alpha\) is the coefficient of linear expansion. (b) A circular sheet of aluminum is 55.0 \(\mathrm{cm}\) in diameter at \(15.0^{\circ} \mathrm{C} .\) By how much does the area of one side of the sheet change when the temperature increases to \(27.5^{\circ} \mathrm{C} ?\)

While vacationing in Italy, you see on local TV one summer morning that temperature will rise from the current \(18^{\circ} \mathrm{C}\) to a high of \(39^{\circ} \mathrm{C}\) . What is the corresponding increase in the Fahrenheit temperature?

Bulk Stress Due to a Temperature Increase. (a) Prove that, if an object under pressure has its temperature raised but is not allowed to expand, the increase in pressure is $$\Delta p=B \beta \Delta T$$ where the bulk modulus \(B\) and the average coefficient of volume expansion \(\beta\) are both assumed positive and constant. (b) What pressure is necessary to prevent a steel block from expanding when its temperature is increased from \(20.0^{\circ} \mathrm{C}\) to \(35.0^{\circ} \mathrm{C} ?\)

A copper calorimeter can with mass 0.100 kg contains 0.160 \(\mathrm{kg}\) of water and 0.0180 \(\mathrm{kg}\) of ice in thermal equilibrium at atmospheric pressure. If 0.750 \(\mathrm{kg}\) of lead at a temperature of \(255^{\circ} \mathrm{C}\) is dropped into the calorimeter can, what is the final temperature? Assume that no heat is lost to the surroundings.
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Electromagnetism

Read Explanation

Mechanics and Materials

Read Explanation

Waves Physics

Read Explanation

Astrophysics

Read Explanation

Physics of Motion

Read Explanation
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.