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A sample of an unknown substance has a mass of 0.465 kg. If 3,000.0 J of heat is required to heat the substance from 50.0°C to 100.0°C, what is the specific heat of the substance?

Use [tex]\( q = m C_p \Delta T \)[/tex].

A. [tex]\( 0.00775 \, \frac{J}{g \, ^\circ C} \)[/tex]
B. [tex]\( 0.0600 \, \frac{J}{g \, ^\circ C} \)[/tex]
C. [tex]\( 0.129 \, \frac{J}{g \, ^\circ C} \)[/tex]
D. [tex]\( 0.155 \, \frac{J}{g \, ^\circ C} \)[/tex]



Answer :

GinnyAnswer
To find the specific heat capacity ([tex]\(C_p\)[/tex]) of the unknown substance, we can use the given formula:

[tex]\[ q = m \cdot C_p \cdot \Delta T \][/tex]

Where:
- [tex]\(q\)[/tex] is the heat added (in joules).
- [tex]\(m\)[/tex] is the mass of the substance (in grams for the specific heat capacity's units to match).
- [tex]\(C_p\)[/tex] is the specific heat capacity (in [tex]\( \frac{J}{g \, ^\circ C} \)[/tex]).
- [tex]\(\Delta T\)[/tex] is the change in temperature (in Celsius).

Let's break this problem down step-by-step.

1. Given data:
- Heat ([tex]\(q\)[/tex]) = 3000.0 J
- Mass ([tex]\(m\)[/tex]) = 0.465 kg
- Initial Temperature = 50.0°C
- Final Temperature = 100.0°C

2. Convert the mass from kilograms to grams:
[tex]\[ m = 0.465 \text{ kg} = 0.465 \times 1000 \text{ g} = 465.0 \text{ g} \][/tex]

3. Calculate the temperature change ([tex]\(\Delta T\)[/tex]):
[tex]\[ \Delta T = \text{Final Temperature} - \text{Initial Temperature} \][/tex]
[tex]\[ \Delta T = 100.0^\circ C - 50.0^\circ C = 50.0^\circ C \][/tex]

4. Rearrange the specific heat capacity formula to solve for [tex]\(C_p\)[/tex]:
[tex]\[ C_p = \frac{q}{m \cdot \Delta T} \][/tex]

5. Substitute the values into the rearranged equation:
[tex]\[ C_p = \frac{3000.0 \text{ J}}{465.0 \text{ g} \times 50.0^\circ C} \][/tex]

6. Calculate [tex]\(C_p\)[/tex]:
[tex]\[ C_p = \frac{3000.0}{465.0 \times 50.0} \][/tex]

7. Simplify the denominator:
[tex]\[ 465.0 \times 50.0 = 23250.0 \][/tex]

8. Finish the calculation:
[tex]\[ C_p = \frac{3000.0}{23250.0} \approx 0.129 \frac{J}{g \, ^\circ C} \][/tex]

The specific heat capacity of the substance is approximately [tex]\( 0.129 \, \frac{J}{g \, ^\circ C} \)[/tex].

Thus, among the given options, the correct specific heat capacity is:

[tex]\[ 0.129 \frac{J}{g \, ^\circ C} \][/tex]

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