/*! 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 23 An aluminum tea kettle with mass... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An aluminum tea kettle with mass 1.10 kg and containing 1.80 kg of water is placed on a stove. If no heat is lost to the surroundings, how much heat must be added to raise the temperature from 20.0\(^\circ\)C to 85.0\(^\circ\)C?

Short Answer

Expert verified
553581 J

Step by step solution

01

Understand the Specific Heat Capacity Equation

The heat required to raise the temperature of an object is given by the formula \( Q = mc\Delta T \), where \( Q \) is the heat added, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.
02

Collect Known Values

The mass of the aluminum kettle \( m_1 = 1.10 \) kg and its specific heat capacity \( c_1 = 900 \) J/(kg·°C). The mass of the water \( m_2 = 1.80 \) kg and its specific heat capacity \( c_2 = 4186 \) J/(kg·°C). The initial temperature \( T_i = 20.0 \)°C, and the final temperature \( T_f = 85.0 \)°C.
03

Calculate Change in Temperature

The change in temperature \( \Delta T = T_f - T_i = 85.0 \text{°C} - 20.0 \text{°C} = 65.0 \text{°C} \).
04

Calculate Heat for Aluminum Kettle

For the aluminum kettle, the heat \( Q_1 = m_1c_1\Delta T = 1.10 \times 900 \times 65.0 = 64350 \text{ J} \).
05

Calculate Heat for Water

For the water, the heat \( Q_2 = m_2c_2\Delta T = 1.80 \times 4186 \times 65.0 = 489231 \text{ J} \).
06

Total Heat Required

The total heat required \( Q = Q_1 + Q_2 = 64350 + 489231 = 553581 \text{ J} \).

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.

Aluminum Heat Calculation
Aluminum is a commonly used metal due to its light weight and good thermal conductivity. When we want to heat aluminum, such as an aluminum tea kettle, we need to determine how much heat is needed to reach our desired temperature. This depends on its specific heat capacity, which is a measure of how much heat energy it takes to raise the temperature of a unit mass of a substance by one degree Celsius.
In our example, the specific heat capacity of aluminum is 900 J/(kg·°C). To calculate the heat required for the aluminum kettle, we use the equation:
  • \( Q = mc\Delta T \)
Where:
  • \( Q \) is the heat energy
  • \( m \) is the mass of the aluminum (1.10 kg)
  • \( c \) is the specific heat capacity (900 J/(kg·°C))
  • \( \Delta T \) is the change in temperature
Inserting the values into the equation, we find:
  • \( Q_1 = 1.10 \times 900 \times 65 = 64350 \) J
This calculation shows us that 64,350 Joules of energy are needed to heat the aluminum kettle from 20.0 °C to 85.0 °C.
Water Heat Calculation
Water has a higher specific heat capacity compared to many other substances, meaning it takes more energy to change its temperature. This property makes water an excellent choice for thermal applications, but it also requires more energy to heat up. The specific heat capacity of water is 4186 J/(kg·°C).
To calculate the heat required to raise the water's temperature, we use the same equation for heat energy:
  • \( Q = mc\Delta T \)
For the water in our kettle, we plug in:
  • \( m = 1.80 \) kg
  • \( c = 4186 \) J/(kg·°C)
  • \( \Delta T = 65.0 \) °C
Calculating this, we get:
  • \( Q_2 = 1.80 \times 4186 \times 65 = 489231 \) J
Thus, 489,231 Joules of energy are required to raise the temperature of the water from 20.0 °C to 85.0 °C.
Temperature Change Calculation
Understanding the concept of temperature change is essential when dealing with thermal energy calculations. Temperature change is calculated as the difference between the final temperature and the initial temperature of a substance. This is represented by \( \Delta T \) in the heat capacity formula, where \( \Delta T = T_f - T_i \).
In our scenario, the temperature change for both the aluminum kettle and the water is the same:
  • Initial temperature, \( T_i = 20.0 \)°C
  • Final temperature, \( T_f = 85.0 \)°C
Using these values, we calculate the temperature change:
  • \( \Delta T = 85.0 - 20.0 = 65.0 \)°C
By knowing \( \Delta T \), we can accurately determine the amount of heat energy required to raise the temperature for both the aluminum kettle and the water. This consistency in temperature change across different materials simplifies energy calculations because it only needs to be computed once for all components heated together.

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

CP A 15.0-g bullet traveling horizontally at 865 m/s passes through a tank containing 13.5 kg of water and emerges with a speed of 534 m/s. What is the maximum temperature increase that the water could have as a result of this event?

BIO While running, a 70-kg student generates thermal energy at a rate of 1200 W. For the runner to maintain a constant body temperature of 37\(^\circ\)C, this energy must be removed by perspiration or other mechanisms. If these mechanisms failed and the energy could not flow out of the student’s body, for what amount of time could a student run before irreversible body damage occurred? (Note: Protein structures in the body are irreversibly damaged if body temperature rises to 44\(^\circ\)C or higher. The specific heat of a typical human body is 3480 J / kg \(\cdot\) K, slightly less than that of water. The difference is due to the presence of protein, fat, and minerals, which have lower specific heats.)

What is the rate of energy radiation per unit area of a blackbody at (a) 273 K and (b) 2730 K?

Like the Kelvin scale, the Rankine scale is an absolute temperature scale: Absolute zero is zero degrees Rankine (0\(^\circ\)R). However, the units of this scale are the same size as those of the Fahrenheit scale rather than the Celsius scale. What is the numerical value of the triple-point temperature of water on the Rankine scale?

A long rod, insulated to prevent heat loss along its sides, is in perfect thermal contact with boiling water (at atmospheric pressure) at one end and with an ice-water mixture at the other (Fig. E17.62). The rod consists of a 1.00-m section of copper (one end in boiling water) joined end to end to a length \(L_2\) of steel (one end in the ice-water mixture). Both sections of the rod have crosssectional areas of 4.00 cm\(^2\). The temperature of the copper- steel junction is 65.0\(^\circ\)C after a steady state has been set up. (a) How much heat per second flows from the boiling water to the ice-water mixture? (b) What is the length \(L_2\) of the steel section?

See all solutions

Recommended explanations on Physics 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.