Watch KCV: The First Law of Thermodynamics. Review Temperature Changes and Heat Capacity; Read Section 7.3.

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

Part C: Ethanol ([tex]$T_f = 44.4^{\circ}C$[/tex])
Express your answer in grams to three significant figures.
[tex]$m = \square$[/tex] g
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Part D: Water ([tex]$T_f = 32.4^{\circ}C$[/tex])
Express your answer in grams to three significant figures.
[tex]$m = \square$[/tex] g
Submit
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Answer :

### Part C - Ethanol
We start with the initial temperature of the ethanol, which is [tex]\(25.0^\circ C\)[/tex], and it absorbs 1.98 × 10³ J of heat. The final temperature recorded for ethanol is [tex]\(44.4^\circ C\)[/tex].

1. Temperature Change [tex]\( (\Delta T) \)[/tex]:
- Initial temperature [tex]\( T_i = 25.0^\circ C \)[/tex]
- Final temperature [tex]\( T_f = 44.4^\circ C \)[/tex]
- Change in temperature [tex]\( \Delta T = T_f - T_i = 44.4^\circ C - 25.0^\circ C = 19.4^\circ C \)[/tex]

2. Calculate the mass [tex]\( m \)[/tex] using the formula [tex]\( Q = mc\Delta T \)[/tex]:
- [tex]\( Q \)[/tex]: Heat absorbed = 1.98 × 10³ J
- [tex]\( c \)[/tex]: Specific heat capacity of ethanol = 2.42 J/(g·°C)
- Rearrange the formula to solve for mass [tex]\( m \)[/tex]:
[tex]\[ m = \frac{Q}{c \Delta T} = \frac{1.98 \times 10^3 \, J}{2.42 \, J/(g \cdot ° C) \times 19.4^\circ C} \][/tex]

3. Mass Calculation:
[tex]\[ m \approx 42.174 \, g \][/tex]

Therefore, the mass of the ethanol is [tex]\( \boxed{42.174} \)[/tex] grams.

### Part D - Water
Now, we consider the initial temperature of the water, which is also [tex]\(25.0^\circ C\)[/tex], and it absorbs 1.98 × 10³ J of heat. The final temperature recorded for water is [tex]\(32.4^\circ C\)[/tex].

1. Temperature Change [tex]\( (\Delta T) \)[/tex]:
- Initial temperature [tex]\( T_i = 25.0^\circ C \)[/tex]
- Final temperature [tex]\( T_f = 32.4^\circ C \)[/tex]
- Change in temperature [tex]\( \Delta T = T_f - T_i = 32.4^\circ C - 25.0^\circ C = 7.4^\circ C \)[/tex]

2. Calculate the mass [tex]\( m \)[/tex] using the formula [tex]\( Q = mc\Delta T \)[/tex]:
- [tex]\( Q \)[/tex]: Heat absorbed = 1.98 × 10³ J
- [tex]\( c \)[/tex]: Specific heat capacity of water = 4.18 J/(g·°C)
- Rearrange the formula to solve for mass [tex]\( m \)[/tex]:
[tex]\[ m = \frac{Q}{c \Delta T} = \frac{1.98 \times 10^3 \, J}{4.18 \, J/(g \cdot ° C) \times 7.4^\circ C} \][/tex]

3. Mass Calculation:
[tex]\[ m \approx 64.011 \, g \][/tex]

Therefore, the mass of the water is [tex]\( \boxed{64.011} \)[/tex] grams.