madey21
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The overall heat capacity of the calorimeter is [tex]$8.67 \, \text{kJ/}^{\circ} \text{C}$[/tex]. If the calorimeter contained [tex]$1.500 \, \text{kg}$[/tex] of water, what is the heat capacity of the dry portion of the calorimeter?

[tex]\[
\begin{array}{r}
C_{\text{cal (complete)}} = C_{\text{cal (dry)}} + C_{\text{cal} \left( \text{H}_2 \text{O} \right)} \\
C_{\text{H}_2 \text{O}} = 4.184 \, \text{J/g}^{\circ} \text{C} \\
8.67 \, \text{kJ/}^{\circ} \text{C} = C_{\text{cal (dry)}} + \left( 1.500 \, \text{kg} \times 4.184 \, \text{kJ/kg}^{\circ} \text{C} \right)
\end{array}
\][/tex]



Answer :

GinnyAnswer
Sure, let's go through the solution step-by-step. We want to find the heat capacity of the dry portion of the calorimeter. We are given the overall heat capacity of the calorimeter with water and the heat capacity of water.

Here are the steps needed to solve this problem:

1. Identify the given data:
- Overall heat capacity of the calorimeter (complete calorimeter with water): [tex]\( C_{\text{cal (complete)}} = 8.67 \ \text{kJ}/{^\circ}C \)[/tex]
- Mass of the water: [tex]\( m_{\text{water}} = 1.500 \ \text{kg} \)[/tex]
- Specific heat capacity of water: [tex]\( c_{\text{water}} = 4.184 \ \text{kJ}/\text{kg} {^\circ}C \)[/tex]

2. Calculate the heat capacity of the water:
The heat capacity of water can be calculated using the formula:
[tex]\[ C_{\text{water}} = m_{\text{water}} \times c_{\text{water}} \][/tex]
Plugging in the given values:
[tex]\[ C_{\text{water}} = 1.500 \ \text{kg} \times 4.184 \ \text{kJ}/\text{kg} {^\circ}C \][/tex]
[tex]\[ C_{\text{water}} = 6.276 \ \text{kJ}/{^\circ}C \][/tex]

3. Relate the heat capacities to find the dry portion:
We know that the total heat capacity is the sum of the heat capacity of the dry portion of the calorimeter and the heat capacity of water.
[tex]\[ C_{\text{cal (complete)}} = C_{\text{cal (dry)}} + C_{\text{water}} \][/tex]
Rearrange this equation to solve for [tex]\( C_{\text{cal (dry)}} \)[/tex]:
[tex]\[ C_{\text{cal (dry)}} = C_{\text{cal (complete)}} - C_{\text{water}} \][/tex]
Substitute the known values:
[tex]\[ C_{\text{cal (dry)}} = 8.67 \ \text{kJ}/{^\circ}C - 6.276 \ \text{kJ}/{^\circ}C \][/tex]
[tex]\[ C_{\text{cal (dry)}} = 2.394 \ \text{kJ}/{^\circ}C \][/tex]

So, the heat capacity of the dry portion of the calorimeter is [tex]\( 2.394 \ \text{kJ}/{^\circ}C \)[/tex], and the heat capacity of the water is [tex]\( 6.276 \ \text{kJ}/{^\circ}C \)[/tex].

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