madey21
Answered

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During the reaction, the water temperature rises from [tex]$25.00^{\circ}C$[/tex] to [tex]$28.54^{\circ}C$[/tex].

The heat capacity of the calorimeter, including water, is [tex]$5860 \, \text{J}/^{\circ}\text{C}$[/tex]. Calculate the heat of reaction in Joules.

[tex]q_{\text{rxn}} = \, ? \, \text{J}[/tex]

Hint: Let the units guide your calculation setup! Enter either a + or - sign and the magnitude. Do not round until the end.


Sagot :

Sure, let’s walk through the problem step by step.

### Step 1: Identify the initial and final temperatures
Given:
- Initial temperature ([tex]\( T_{\text{initial}} \)[/tex]): [tex]\( 25.00 ^{\circ} \text{C} \)[/tex]
- Final temperature ([tex]\( T_{\text{final}} \)[/tex]): [tex]\( 28.54 ^{\circ} \text{C} \)[/tex]

### Step 2: Calculate the change in temperature
The change in temperature ([tex]\( \Delta T \)[/tex]) is calculated as the difference between the final and initial temperatures:

[tex]\[ \Delta T = T_{\text{final}} - T_{\text{initial}} \][/tex]

Substituting the given values:

[tex]\[ \Delta T = 28.54 ^{\circ} \text{C} - 25.00 ^{\circ} \text{C} = 3.54 ^{\circ} \text{C} \][/tex]

The exact value (without rounding) is:

[tex]\[ \Delta T = 3.539999999999999 ^{\circ} \text{C} \][/tex]

### Step 3: Identify the heat capacity of the calorimeter
Given:
- Heat capacity ([tex]\( C \)[/tex]): [tex]\( 5860 \, \text{J/}^{\circ} \text{C} \)[/tex]

### Step 4: Calculate the heat of reaction
The heat of reaction [tex]\( q_{\text{rxn}} \)[/tex] is given by the formula:

[tex]\[ q_{\text{rxn}} = C \times \Delta T \][/tex]

Substitute the known values:

[tex]\[ q_{\text{rxn}} = 5860 \, \text{J/}^{\circ} \text{C} \times 3.539999999999999 ^{\circ} \text{C} \][/tex]

Performing the multiplication:

[tex]\[ q_{\text{rxn}} = 5860 \times 3.539999999999999 = 20744.399999999994 \, \text{J} \][/tex]

### Final Answer:
The heat of reaction is:

[tex]\[ q_{\text{rxn}} = 20744.399999999994 \, \text{J} \][/tex]

Thus, during the reaction, the heat released (or absorbed) by the calorimeter and the water is approximately [tex]\( 20744.40 \, \text{J} \)[/tex] when rounded to two decimal places. If you do not require rounding, you can use the value [tex]\( 20744.399999999994 \, \text{J} \)[/tex].
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