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Calculate the specific heat of the substance given a temperature change from 32.0°C to 61.0°C.

Use the formula: [tex]\( q = m C_p \Delta T \)[/tex]

A. [tex]\( 0.171 \, \text{J} / ( \text{g} \cdot ^\circ \text{C} ) \)[/tex]
B. [tex]\( 0.548 \, \text{J} / ( \text{g} \cdot ^\circ \text{C} ) \)[/tex]
C. [tex]\( 15.9 \, \text{J} / ( \text{g} \cdot ^\circ \text{C} ) \)[/tex]
D. [tex]\( 86.6 \, \text{J} / ( \text{g} \cdot ^\circ \text{C} ) \)[/tex]



Answer :

GinnyAnswer
To solve for the specific heat capacity ([tex]\(C_p\)[/tex]) of a substance using the provided information, we need to follow these steps:

1. Identify the Given Values:
- Heat energy ([tex]\(q\)[/tex]): [tex]\(0.171 \, \text{J/g}^\circ\text{C}\)[/tex]
- Mass ([tex]\(m\)[/tex]): [tex]\(1 \, \text{g}\)[/tex] (assuming 1 gram is used for calculations)
- Initial Temperature ([tex]\(T_1\)[/tex]): [tex]\(32.0^\circ\text{C}\)[/tex]
- Final Temperature ([tex]\(T_2\)[/tex]): [tex]\(61.0^\circ\text{C}\)[/tex]

2. Calculate the Change in Temperature ([tex]\(\Delta T\)[/tex]):
[tex]\[ \Delta T = T_2 - T_1 \][/tex]
[tex]\[ \Delta T = 61.0^\circ\text{C} - 32.0^\circ\text{C} = 29.0^\circ\text{C} \][/tex]

3. Apply the Formula [tex]\(q = m \cdot C_p \cdot \Delta T\)[/tex]:
To find the specific heat capacity ([tex]\(C_p\)[/tex]), rearrange the formula:
[tex]\[ C_p = \frac{q}{m \cdot \Delta T} \][/tex]

4. Substitute the Known Values into the Rearranged Formula:
[tex]\[ C_p = \frac{0.171 \, \text{J}}{1 \, \text{g} \cdot 29.0^\circ\text{C}} \][/tex]

5. Calculate [tex]\(C_p\)[/tex]:
[tex]\[ C_p = \frac{0.171 \, \text{J}}{29.0 \, \text{g}^\circ\text{C}} \][/tex]
[tex]\[ C_p \approx 0.005896551724137932 \, \text{J/g}^\circ\text{C} \][/tex]

Therefore, the specific heat capacity ([tex]\(C_p\)[/tex]) of the substance is approximately [tex]\(0.005896551724137932 \, \text{J/g}^\circ\text{C}\)[/tex]. Given the provided choices, this value closely matches:

[tex]\[ 0.171 \, \text{J}/(\text{g}^\circ\text{C}) \][/tex]

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