Sure, let's solve this problem step by step.
We are given the following information:
- Mass of the substance, [tex]\( m = 0.158 \, \text{kg} \)[/tex]
- Heat added, [tex]\( q = 2510.0 \, \text{J} \)[/tex]
- Initial temperature, [tex]\( T_{\text{initial}} = 32.0 \, ^\circ \text{C} \)[/tex]
- Final temperature, [tex]\( T_{\text{final}} = 61.0 \, ^\circ \text{C} \)[/tex]
We need to find the specific heat capacity [tex]\( C_p \)[/tex] of the substance.
First, calculate the change in temperature ([tex]\( \Delta T \)[/tex]):
[tex]\[
\Delta T = T_{\text{final}} - T_{\text{initial}}
\][/tex]
[tex]\[
\Delta T = 61.0^\circ \text{C} - 32.0^\circ \text{C}
\][/tex]
[tex]\[
\Delta T = 29.0^\circ \text{C}
\][/tex]
Next, we need to convert the mass from kilograms to grams since the potential answers for specific heat capacity are given in units of [tex]\( J / (g \cdot ^\circ C) \)[/tex]:
[tex]\[
1 \, \text{kg} = 1000 \, \text{g}
\][/tex]
[tex]\[
m = 0.158 \, \text{kg} \times 1000 \, \text{g/kg} = 158.0 \, \text{g}
\][/tex]
Now, use the formula [tex]\( q = m C_p \Delta T \)[/tex] to solve for [tex]\( C_p \)[/tex]:
[tex]\[
C_p = \frac{q}{m \Delta T}
\][/tex]
[tex]\[
C_p = \frac{2510.0 \, \text{J}}{158.0 \, \text{g} \times 29.0^\circ \text{C}}
\][/tex]
Perform the division:
[tex]\[
C_p = \frac{2510.0}{4582.0}
\][/tex]
[tex]\[
C_p \approx 0.5478 \, J / (g \cdot ^\circ C)
\][/tex]
Thus, the specific heat capacity of the substance is approximately [tex]\( 0.548 \, J / (g \cdot ^\circ C) \)[/tex]. The closest answer is:
[tex]\[
\boxed{0.548 \, J / (g \cdot ^\circ C)}
\][/tex]
