/*! 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 11 The initial temperature of a 344... [FREE SOLUTION] | 91Ó°ÊÓ

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The initial temperature of a 344 -g sample of iron is \(18.2^{\circ} \mathrm{C} .\) If the sample absorbs \(2.25 \mathrm{kJ}\) of heat, what is its final temperature?

Short Answer

Expert verified
The final temperature of the iron is approximately 32.77°C.

Step by step solution

01

Understand the problem

We need to determine the final temperature of a 344-g sample of iron initially at 18.2°C after it absorbs 2.25 kJ of heat.
02

Identify the formula

We use the formula for calculating heat change, which is \( q = mc\Delta T \), where \( q \) is the heat added, \( m \) is the mass of the substance, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.
03

Gather known values

- \( m = 344 \) g (iron's mass) - \( q = 2.25 \) kJ (heat absorbed, convert this to J: \( 2.25 \times 1000 = 2250 \) J) - \( c = 0.449 \) J/g°C (specific heat capacity of iron) - Initial temperature \( T_i = 18.2^{\circ} \mathrm{C} \) (Given)- Final temperature \( T_f \) (Needs to be calculated)
04

Use the formula to find \( \Delta T \)

Rearrange the formula to solve for \( \Delta T \): \( \Delta T = \frac{q}{mc} \) Substitute the known values: \( \Delta T = \frac{2250}{344 \times 0.449} \) \( \Delta T = \frac{2250}{154.456} \) \( \Delta T \approx 14.57^{\circ} \mathrm{C} \) (rounded to two decimal places).
05

Calculate the final temperature \( T_f \)

To find the final temperature, use the formula: \( T_f = T_i + \Delta T \) Substitute the known values: \( T_f = 18.2 + 14.57 \) \( T_f \approx 32.77^{\circ} \mathrm{C} \) (rounded to two decimal places).

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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 physics that describes the movement of energy from one object to another because of a temperature difference. When heat flows into or out of an object, it causes a change in the energy contained within that object. In our exercise, the 344-g sample of iron absorbs 2.25 kJ of heat, illustrating the process of heat transfer.
There are three main modes of heat transfer:
  • Conduction: Heat transfer through direct contact between materials, like a metal spoon getting hot when placed in a pot of boiling water.
  • Convection: Heat transfer through fluid motion, such as warm air rising and cool air descending in a room.
  • Radiation: Heat transfer via electromagnetic waves, like the warmth from sunlight.
In this specific exercise, heat is transferred to iron by conduction, which results in an increase in its thermal energy. Understanding how heat moves and the effects of heat transfer is crucial in solving problems like the one posed by the exercise.
Temperature Change
The change in temperature (\(\Delta T\)) occurs when an object either gains or loses heat energy. Specifically, in the context of our exercise, the temperature of the iron increases as it absorbs heat. The relationship between heat transfer and temperature change can be calculated using the formula: \( q = mc\Delta T \),where:
  • \(q\) is the heat energy added or removed,
  • \(m\) is the mass,
  • \(c\) is the specific heat capacity,
  • \(\Delta T\) is the temperature change.
In our solution, once we have all the known values: the iron mass (\(344\) g), the specific heat capacity (\(0.449\) J/g°C), and the heat added (\(2250\) J), we can determine how much the temperature of iron will change. The calculated \(\Delta T\) of approximately \(14.57^{\circ} \mathrm{C}\) shows how much hotter the iron became due to the heat absorbed.
Calorimetry
Calorimetry is the science of measuring the amount of heat involved in a chemical or physical process. It can be used to measure changes in temperature, phase changes, or chemical reactions. In this exercise, calorimetry allows us to calculate the final temperature of the iron sample after absorbing heat.In essence, calorimetry involves the use of a calorimeter to compute the heat exchanged during physical changes like melting or heating. By knowing the specific heat capacity, mass, and amount of heat added, we can find out how much the temperature will change in the substance.
Components in calorimetry:
  • **Calorimeter:** A device used to measure temperature changes.
  • **Temperature measurements:** Initial and final temperature readings are key to finding heat change.
  • **Specific Heat Capacity (\(c\):**A crucial value that indicates how much heat is needed to change the temperature of a unit mass by 1°C.
Understanding calorimetry and its equations allows us to solve the problem effectively, giving insight into how much and how fast heat can change the temperature of different substances, like in our iron sample example.

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Most popular questions from this chapter

You are on a diet that calls for eating no more than 1200 Cal/day. How many joules would this be?

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