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Read Section 7.3 on Temperature Changes and Heat Capacity. You can click on the Review link to access the section in your eText.

An unknown mass of each substance, initially at [tex]$25.0^{\circ} C$[/tex], absorbs [tex]$1.98 \times 10^3 J$[/tex] of heat. The final temperature is recorded. Find the mass of each substance. Specific heat capacities are given in the table below.

\begin{tabular}{|l|l|}
\hline
Substance & \begin{tabular}{l}
Specific Heat Capacity at \\
[tex][tex]$298 K, C_{s}\left(J / g \cdot { }^{\circ}C\right)$[/tex]
\end{tabular} \\
\hline
\begin{tabular}{l}
Glass \\
(Pyrex)
\end{tabular} & 0.75 \\
\hline
Sand & 0.84 \\
\hline
Ethanol & 2.42 \\
\hline
Water & 4.18 \\
\hline
\end{tabular}

Part A: Pyrex Glass
Final temperature: [tex]$T_{f} = 55.3^{\circ} C$[/tex]
Express your answer in grams to two significant figures.
[tex]\square[/tex]

Part B: Sand
Final temperature: [tex]$T_{f} = 62.0^{\circ} C$[/tex]
Express your answer in grams to two significant figures.
[tex]\square[/tex]

Sagot :

To find the masses of both substances (Pyrex glass and sand), we'll follow a systematic approach making use of the heat absorbed, the specific heat capacities, and the temperature changes.

### Part A: Mass of Pyrex Glass

1. Given Data:
- Heat absorbed ([tex]\(q\)[/tex]) = [tex]\(1.98 \times 10^3 \text{ J}\)[/tex]
- Initial temperature ([tex]\(T_i\)[/tex]) = [tex]\(25.0^{\circ} \text{C}\)[/tex]
- Final temperature for glass ([tex]\(T_f\)[/tex]) = [tex]\(55.3^{\circ} \text{C}\)[/tex]
- Specific heat capacity of Pyrex glass ([tex]\(c\)[/tex]) = [tex]\(0.75 \text{ J/g} \cdot ^{\circ} \text{C}\)[/tex]

2. Temperature Change ([tex]\(\Delta T\)[/tex]):
[tex]\[ \Delta T_{\text{glass}} = T_f - T_i = 55.3^{\circ} \text{C} - 25.0^{\circ} \text{C} = 30.3^{\circ} \text{C} \][/tex]

3. Using the Heat Formula:
The formula to relate heat absorbed to mass, specific heat capacity, and temperature change is:
[tex]\[ q = m \cdot c \cdot \Delta T \][/tex]
Rearranging the formula to solve for mass ([tex]\(m\)[/tex]):
[tex]\[ m = \frac{q}{c \cdot \Delta T} \][/tex]

4. Calculating the Mass:
[tex]\[ m_{\text{glass}} = \frac{1.98 \times 10^3 \text{ J}}{0.75 \text{ J/g} \cdot ^{\circ} \text{C} \times 30.3^{\circ} \text{C}} = 87.13 \text{ g} \][/tex]

5. Final Answer:
The mass of the Pyrex glass is [tex]\(87.13 \text{ g}\)[/tex].

### Part B: Mass of Sand

1. Given Data:
- Heat absorbed ([tex]\(q\)[/tex]) = [tex]\(1.98 \times 10^3 \text{ J}\)[/tex]
- Initial temperature ([tex]\(T_i\)[/tex]) = [tex]\(25.0^{\circ} \text{C}\)[/tex]
- Final temperature for sand ([tex]\(T_f\)[/tex]) = [tex]\(62.0^{\circ} \text{C}\)[/tex]
- Specific heat capacity of sand ([tex]\(c\)[/tex]) = [tex]\(0.84 \text{ J/g} \cdot ^{\circ} \text{C}\)[/tex]

2. Temperature Change ([tex]\(\Delta T\)[/tex]):
[tex]\[ \Delta T_{\text{sand}} = T_f - T_i = 62.0^{\circ} \text{C} - 25.0^{\circ} \text{C} = 37.0^{\circ} \text{C} \][/tex]

3. Using the Heat Formula:
Rearranging the heat formula to solve for mass ([tex]\(m\)[/tex]):
[tex]\[ m = \frac{q}{c \cdot \Delta T} \][/tex]

4. Calculating the Mass:
[tex]\[ m_{\text{sand}} = \frac{1.98 \times 10^3 \text{ J}}{0.84 \text{ J/g} \cdot ^{\circ} \text{C} \times 37.0^{\circ} \text{C}} = 63.71 \text{ g} \][/tex]

5. Final Answer:
The mass of the sand is [tex]\(63.71 \text{ g}\)[/tex].

This process allows us to determine the mass of both Pyrex glass and sand given the heat absorbed and the specific heat capacities.