Question

calculate the temperature change that results from adding 250 j of thermal energy to each of the following (a) 0.75 mol of hg; (b) 0.35 mol of hg; (c) 0.35 mol of h₂o.
\begin{array}{|l|r|} hline & c_m , (\text{j mol}^{-1} , ^circ\text{c}^{-1}) \\ hline \text{hg}(l) & 27.983 \\ hline \text{h}_2\text{o}(l) & 75.291 \\ hline end{array}

Explanation:

To solve for the temperature change (\(\Delta T\)) when thermal energy (\(q\)) is added, we use the formula for molar heat capacity:

$$ q = n \cdot C_m \cdot \Delta T $$

Rearranging to solve for \(\Delta T\):

$$ \Delta T = \frac{q}{n \cdot C_m} $$

Part (a): 0.75 mol of Hg

Given:

• \( q = 250 \, \text{J} \)
• \( n = 0.75 \, \text{mol} \)
• \( C_m (\text{Hg}) = 27.983 \, \text{J mol}^{-1} \text{°C}^{-1} \)

Step 1: Substitute values into \(\Delta T\) formula

$$ \Delta T = \frac{250 \, \text{J}}{(0.75 \, \text{mol}) \cdot (27.983 \, \text{J mol}^{-1} \text{°C}^{-1})} $$

Step 2: Calculate the denominator

\( 0.75 \cdot 27.983 \approx 20.987 \, \text{J °C}^{-1} \)

Step 3: Solve for \(\Delta T\)

$$ \Delta T \approx \frac{250}{20.987} \approx 11.9 \, \text{°C} $$

Part (b): 0.35 mol of Hg

Given:

• \( q = 250 \, \text{J} \)
• \( n = 0.35 \, \text{mol} \)
• \( C_m (\text{Hg}) = 27.983 \, \text{J mol}^{-1} \text{°C}^{-1} \)

Step 1: Substitute values into \(\Delta T\) formula

$$ \Delta T = \frac{250 \, \text{J}}{(0.35 \, \text{mol}) \cdot (27.983 \, \text{J mol}^{-1} \text{°C}^{-1})} $$

Step 2: Calculate the denominator

\( 0.35 \cdot 27.983 \approx 9.794 \, \text{J °C}^{-1} \)

Step 3: Solve for \(\Delta T\)

$$ \Delta T \approx \frac{250}{9.794} \approx 25.5 \, \text{°C} $$

Part (c): 0.35 mol of \(\boldsymbol{\text{H}_2\text{O}}\)

Given:

• \( q = 250 \, \text{J} \)
• \( n = 0.35 \, \text{mol} \)
• \( C_m (\text{H}_2\text{O}) = 75.291 \, \text{J mol}^{-1} \text{°C}^{-1} \)

Step 1: Substitute values into \(\Delta T\) formula

$$ \Delta T = \frac{250 \, \text{J}}{(0.35 \, \text{mol}) \cdot (75.291 \, \text{J mol}^{-1} \text{°C}^{-1})} $$

Step 2: Calculate the denominator

\( 0.35 \cdot 75.291 \approx 26.352 \, \text{J °C}^{-1} \)

Step 3: Solve for \(\Delta T\)

$$ \Delta T \approx \frac{250}{26.352} \approx 9.49 \, \text{°C} $$

Final Answers:

(a) \(\boldsymbol{\approx 11.9 \, \text{°C}}\)
(b) \(\boldsymbol{\approx 25.5 \, \text{°C}}\)
(c) \(\boldsymbol{\approx 9.5 \, \text{°C}}\) (rounded to two significant figures)

Answer:

To solve for the temperature change (\(\Delta T\)) when thermal energy (\(q\)) is added, we use the formula for molar heat capacity:

$$ q = n \cdot C_m \cdot \Delta T $$

Rearranging to solve for \(\Delta T\):

$$ \Delta T = \frac{q}{n \cdot C_m} $$

Part (a): 0.75 mol of Hg

Given:

• \( q = 250 \, \text{J} \)
• \( n = 0.75 \, \text{mol} \)
• \( C_m (\text{Hg}) = 27.983 \, \text{J mol}^{-1} \text{°C}^{-1} \)

Step 1: Substitute values into \(\Delta T\) formula

$$ \Delta T = \frac{250 \, \text{J}}{(0.75 \, \text{mol}) \cdot (27.983 \, \text{J mol}^{-1} \text{°C}^{-1})} $$

Step 2: Calculate the denominator

\( 0.75 \cdot 27.983 \approx 20.987 \, \text{J °C}^{-1} \)

Step 3: Solve for \(\Delta T\)

$$ \Delta T \approx \frac{250}{20.987} \approx 11.9 \, \text{°C} $$

Part (b): 0.35 mol of Hg

Given:

• \( q = 250 \, \text{J} \)
• \( n = 0.35 \, \text{mol} \)
• \( C_m (\text{Hg}) = 27.983 \, \text{J mol}^{-1} \text{°C}^{-1} \)

Step 1: Substitute values into \(\Delta T\) formula

$$ \Delta T = \frac{250 \, \text{J}}{(0.35 \, \text{mol}) \cdot (27.983 \, \text{J mol}^{-1} \text{°C}^{-1})} $$

Step 2: Calculate the denominator

\( 0.35 \cdot 27.983 \approx 9.794 \, \text{J °C}^{-1} \)

Step 3: Solve for \(\Delta T\)

$$ \Delta T \approx \frac{250}{9.794} \approx 25.5 \, \text{°C} $$

Part (c): 0.35 mol of \(\boldsymbol{\text{H}_2\text{O}}\)

Given:

• \( q = 250 \, \text{J} \)
• \( n = 0.35 \, \text{mol} \)
• \( C_m (\text{H}_2\text{O}) = 75.291 \, \text{J mol}^{-1} \text{°C}^{-1} \)

Step 1: Substitute values into \(\Delta T\) formula

$$ \Delta T = \frac{250 \, \text{J}}{(0.35 \, \text{mol}) \cdot (75.291 \, \text{J mol}^{-1} \text{°C}^{-1})} $$

Step 2: Calculate the denominator

\( 0.35 \cdot 75.291 \approx 26.352 \, \text{J °C}^{-1} \)

Step 3: Solve for \(\Delta T\)

$$ \Delta T \approx \frac{250}{26.352} \approx 9.49 \, \text{°C} $$

Final Answers:

(a) \(\boldsymbol{\approx 11.9 \, \text{°C}}\)
(b) \(\boldsymbol{\approx 25.5 \, \text{°C}}\)
(c) \(\boldsymbol{\approx 9.5 \, \text{°C}}\) (rounded to two significant figures)