Question

thermodynamics
3.11 an electric heater supplies 25.0 joules of energy to each of the following samples. compute the final temperature in each case: (a) 10.0 g block of al originally at 15.0 °c; (b) 25.0 g block of al originally at 29.5 °c; (c) 25.0 g block of ag originally at 29.5 °c; and (d) 25.0 g sample of liquid h₂o originally at 22.0 °c.
3.12 each of the following is placed in an ice bath...

Explanation:

To solve for the final temperature in each case, we use the formula for heat transfer: \( q = mc\Delta T \), where \( q \) is the heat energy, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T = T_f - T_i \) (change in temperature, final temperature minus initial temperature). Rearranging for \( T_f \), we get:

\[ T_f = T_i + \frac{q}{mc} \]

We need the specific heat capacities (\( c \)):

• \( c_{\text{Al}} = 0.900 \, \text{J/g·°C} \)
• \( c_{\text{Ag}} = 0.235 \, \text{J/g·°C} \)
• \( c_{\text{H}_2\text{O (liquid)}} = 4.184 \, \text{J/g·°C} \)

Part (a): 10.0 g Al, \( T_i = 15.0 \, ^\circ\text{C} \), \( q = 25.0 \, \text{J} \)

Step 1: Identify values

\( m = 10.0 \, \text{g} \), \( c = 0.900 \, \text{J/g·°C} \), \( T_i = 15.0 \, ^\circ\text{C} \), \( q = 25.0 \, \text{J} \)

Step 2: Calculate \( \frac{q}{mc} \)

\[ \frac{q}{mc} = \frac{25.0 \, \text{J}}{(10.0 \, \text{g})(0.900 \, \text{J/g·°C})} = \frac{25.0}{9.00} \approx 2.78 \, ^\circ\text{C} \]

Step 3: Find \( T_f \)

\[ T_f = 15.0 \, ^\circ\text{C} + 2.78 \, ^\circ\text{C} \approx 17.8 \, ^\circ\text{C} \]

Part (b): 25.0 g Al, \( T_i = 29.5 \, ^\circ\text{C} \), \( q = 25.0 \, \text{J} \)

Step 1: Identify values

\( m = 25.0 \, \text{g} \), \( c = 0.900 \, \text{J/g·°C} \), \( T_i = 29.5 \, ^\circ\text{C} \), \( q = 25.0 \, \text{J} \)

Step 2: Calculate \( \frac{q}{mc} \)

\[ \frac{q}{mc} = \frac{25.0 \, \text{J}}{(25.0 \, \text{g})(0.900 \, \text{J/g·°C})} = \frac{25.0}{22.5} \approx 1.11 \, ^\circ\text{C} \]

Step 3: Find \( T_f \)

\[ T_f = 29.5 \, ^\circ\text{C} + 1.11 \, ^\circ\text{C} \approx 30.6 \, ^\circ\text{C} \]

Part (c): 25.0 g Ag, \( T_i = 29.5 \, ^\circ\text{C} \), \( q = 25.0 \, \text{J} \)

Step 1: Identify values

\( m = 25.0 \, \text{g} \), \( c = 0.235 \, \text{J/g·°C} \), \( T_i = 29.5 \, ^\circ\text{C} \), \( q = 25.0 \, \text{J} \)

Step 2: Calculate \( \frac{q}{mc} \)

\[ \frac{q}{mc} = \frac{25.0 \, \text{J}}{(25.0 \, \text{g})(0.235 \, \text{J/g·°C})} = \frac{25.0}{5.875} \approx 4.26 \, ^\circ\text{C} \]

Step 3: Find \( T_f \)

\[ T_f = 29.5 \, ^\circ\text{C} + 4.26 \, ^\circ\text{C} \approx 33.8 \, ^\circ\text{C} \]

Part (d): 25.0 g \( \text{H}_2\text{O (liquid)} \), \( T_i = 22.0 \, ^\circ\text{C} \), \( q = 25.0 \, \text{J} \)

Step 1: Identify values

\( m = 25.0 \, \text{g} \), \( c = 4.184 \, \text{J/g·°C} \), \( T_i = 22.0 \, ^\circ\text{C} \), \( q = 25.0 \, \text{J} \)

Step 2: Calculate \( \frac{q}{mc} \)

\[ \frac{q}{mc} = \frac{25.0 \, \text{J}}{(25.0 \, \text{g})(4.184 \, \text{J/g·°C})} = \frac{25.0}{104.6} \approx 0.239 \, ^\circ\text{C} \]

Step 3: Find \( T_f \)

\[ T_f = 22.0 \, ^\circ\text{C} + 0.239 \, ^\circ\text{C} \approx 22.2 \, ^\circ\text{C} \]

Answer:

s:
(a) \( \boldsymbol{\approx 17.8 \, ^\circ\text{C}} \)
(b) \( \boldsymbol{\approx 30.6 \, ^\circ\text{C}} \)
(c) \( \boldsymbol{\approx 33.8 \, ^\circ\text{C}} \)
(d) \( \boldsymbol{\approx 22.2 \, ^\circ\text{C}} \)