Answer :
Given the problem, let's break down the question to find the answer:
1. Identify the known values:
- Heat ([tex]\(q\)[/tex]): [tex]\(3000.0 \, J\)[/tex]
- Mass ([tex]\(m\)[/tex]): [tex]\(0.465 \, kg\)[/tex] (which we need to convert to grams, since specific heat is typically in [tex]\(\frac{J}{g \cdot ^\circ C}\)[/tex]). [tex]\(0.465 \, kg = 465.0 \, g\)[/tex]
- Initial temperature ([tex]\(T_1\)[/tex]): [tex]\(50.0^{\circ} C\)[/tex]
- Final temperature ([tex]\(T_2\)[/tex]): [tex]\(100.0^{\circ} C\)[/tex]
2. Calculate the change in temperature ([tex]\(\Delta T\)[/tex]):
- [tex]\(\Delta T = T_2 - T_1 = 100.0^{\circ} C - 50.0^{\circ} C = 50.0^{\circ} C\)[/tex]
3. Use the formula for heat transfer [tex]\(q = m C_p \Delta T\)[/tex] to solve for specific heat ([tex]\(C_p\)[/tex]):
Rearrange the formula to solve for [tex]\(C_p\)[/tex]:
[tex]\[ C_p = \frac{q}{m \Delta T} \][/tex]
4. Substitute the known values into the equation:
[tex]\[ C_p = \frac{3000.0 \, J}{465.0 \, g \times 50.0^{\circ} C} \][/tex]
5. Calculate [tex]\(C_p\)[/tex]:
[tex]\[ C_p = \frac{3000.0}{465.0 \times 50.0} \, \frac{J}{g \cdot ^\circ C} \][/tex]
[tex]\[ C_p = \frac{3000.0}{23250.0} \, \frac{J}{g \cdot ^\circ C} \][/tex]
[tex]\[ C_p \approx 0.129 \, \frac{J}{g \cdot ^\circ C} \][/tex]
Hence, the specific heat of the substance is approximately [tex]\(0.129 \, \frac{J}{g \cdot ^\circ C}\)[/tex].
Considering the multiple choices given:
- [tex]\(0.00775 \, \frac{J}{g \cdot ^\circ C}\)[/tex]
- [tex]\(0.0600 \, \frac{J}{g \cdot ^\circ C}\)[/tex]
- [tex]\(0.129 \, \frac{J}{g \cdot ^\circ C}\)[/tex]
- [tex]\(0.155 \, \frac{J}{g \cdot ^\circ C}\)[/tex]
The correct answer is [tex]\(0.129 \, \frac{J}{g \cdot ^\circ C}\)[/tex].
1. Identify the known values:
- Heat ([tex]\(q\)[/tex]): [tex]\(3000.0 \, J\)[/tex]
- Mass ([tex]\(m\)[/tex]): [tex]\(0.465 \, kg\)[/tex] (which we need to convert to grams, since specific heat is typically in [tex]\(\frac{J}{g \cdot ^\circ C}\)[/tex]). [tex]\(0.465 \, kg = 465.0 \, g\)[/tex]
- Initial temperature ([tex]\(T_1\)[/tex]): [tex]\(50.0^{\circ} C\)[/tex]
- Final temperature ([tex]\(T_2\)[/tex]): [tex]\(100.0^{\circ} C\)[/tex]
2. Calculate the change in temperature ([tex]\(\Delta T\)[/tex]):
- [tex]\(\Delta T = T_2 - T_1 = 100.0^{\circ} C - 50.0^{\circ} C = 50.0^{\circ} C\)[/tex]
3. Use the formula for heat transfer [tex]\(q = m C_p \Delta T\)[/tex] to solve for specific heat ([tex]\(C_p\)[/tex]):
Rearrange the formula to solve for [tex]\(C_p\)[/tex]:
[tex]\[ C_p = \frac{q}{m \Delta T} \][/tex]
4. Substitute the known values into the equation:
[tex]\[ C_p = \frac{3000.0 \, J}{465.0 \, g \times 50.0^{\circ} C} \][/tex]
5. Calculate [tex]\(C_p\)[/tex]:
[tex]\[ C_p = \frac{3000.0}{465.0 \times 50.0} \, \frac{J}{g \cdot ^\circ C} \][/tex]
[tex]\[ C_p = \frac{3000.0}{23250.0} \, \frac{J}{g \cdot ^\circ C} \][/tex]
[tex]\[ C_p \approx 0.129 \, \frac{J}{g \cdot ^\circ C} \][/tex]
Hence, the specific heat of the substance is approximately [tex]\(0.129 \, \frac{J}{g \cdot ^\circ C}\)[/tex].
Considering the multiple choices given:
- [tex]\(0.00775 \, \frac{J}{g \cdot ^\circ C}\)[/tex]
- [tex]\(0.0600 \, \frac{J}{g \cdot ^\circ C}\)[/tex]
- [tex]\(0.129 \, \frac{J}{g \cdot ^\circ C}\)[/tex]
- [tex]\(0.155 \, \frac{J}{g \cdot ^\circ C}\)[/tex]
The correct answer is [tex]\(0.129 \, \frac{J}{g \cdot ^\circ C}\)[/tex].
