**Heat Calculation of an Iron Sample**The temperature of a sample of iron increased by 23.4 °C when 269 J of heat was applied. What is the mass of the sample?\[ m = \underline{\hspace{5cm}} \, \text{g} \]**Specific Heat Table**| Substance | Specific Heat (J/g°C) ||-----------|-----------------------|| Lead | 0.128 || Silver | 0.235 || Copper | 0.385 || Iron | 0.449 || Aluminum | 0.903 |**Explanation of the Diagram:**The table on the right lists substances alongside their specific heat capacities in J/g°C. Specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius. In this problem, we focus on the specific heat of iron to find the mass of the sample using the formula:\[ q = m \cdot c \cdot \Delta T \]Where:- \( q \) is the heat energy (\( 269 \, \text{J} \))- \( m \) is the mass in grams, which we are solving for- \( c \) is the specific heat capacity of iron (\( 0.449 \, \text{J/g°C} \))- \( \Delta T \) is the change in temperature (\( 23.4 \, \text{°C} \))Using these values, you can calculate the mass of the iron sample.

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The temperature of a sample of iron increased by 23.4 °C Substance Specific heat J/(g °C) when 269 J of heat was applied. lead 0.128 What is the mass of the sample? silver 0.235 copper 0.385 iron 0.449 aluminum 0.903 m =

The temperature of a sample of iron increased by 23.4 °C Substance Specific heat J/(g °C) when 269 J of heat was applied. lead 0.128 What is the mass of the sample? silver 0.235 copper 0.385 iron 0.449 aluminum 0.903 m =

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

**Heat Calculation of an Iron Sample**

The temperature of a sample of iron increased by 23.4 °C when 269 J of heat was applied. What is the mass of the sample?

\[ m = \underline{\hspace{5cm}} \, \text{g} \]

**Specific Heat Table**

| Substance | Specific Heat (J/g°C) |
|-----------|-----------------------|
| Lead | 0.128 |
| Silver | 0.235 |
| Copper | 0.385 |
| Iron | 0.449 |
| Aluminum | 0.903 |

**Explanation of the Diagram:**

The table on the right lists substances alongside their specific heat capacities in J/g°C. Specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius. In this problem, we focus on the specific heat of iron to find the mass of the sample using the formula:

\[ q = m \cdot c \cdot \Delta T \]

Where:
- \( q \) is the heat energy (\( 269 \, \text{J} \))
- \( m \) is the mass in grams, which we are solving for
- \( c \) is the specific heat capacity of iron (\( 0.449 \, \text{J/g°C} \))
- \( \Delta T \) is the change in temperature (\( 23.4 \, \text{°C} \))

Using these values, you can calculate the mass of the iron sample.

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