The specific heat capacity, C, of the substance is 3.94 [tex]\frac{J}{g-^oC}[/tex]. To raise one gram of a substance by one degree Celsius, a specific amount of heat must be absorbed. This is called the specific heat of the substance. The higher the specific heat of a substance, the more energy is needed to increase its temperature and vice versa.
Further Explanation:
The specific heat capacity, C, of a substance is related to the amount of substance and the temperature change using the following equation
[tex]q \ = \ mC(T_{f}-T_{i})[/tex]
where
q = amount of heat absorbed (usually in J)
m = amount of substance (in g)
C = specific heat capacity (in [tex]\frac{J}{g-^oC}[/tex])
T(f) = final temperature
T(i) = initial temperature.
To answer the problem, we sort the given information:
q = 22974 J (5 significan figures)
m = 96.5 g (3 significant figures)
C = ?
T(f) = 73.86 °C (4 significant figures)
T(i) = 13.46 °C (4 significant figures)
Then, to find C, rearrange the equation:
[tex]C \ = \frac{q}{m(T_{f} \ - \ T_{i})}[/tex]
Plug in the values to the equation above to get the value of C.
[tex]C \ = \ \frac{22974 \ J}{96.5 \ g \ (73.86 \ \ -13.46 \°C)} \\C \ = \ \frac{22974 \ J}{96.5 \ g \ (60.4\°C)} \\C \ = \ \frac{22974 \ J}{5828.6 \ g -\°C} \\C \ = \ 3.941598 \ \frac{J}{g -\°C} \\[/tex]
Since the least number of significant figures in the given is 3, then the final answer must have 3 significant figures, too. Therefore,
[tex]\boxed {C = 3.94 \ \frac{J}{ g -\°C}}[/tex]
Learn More:
1. Learn more about heat capacity https://brainly.com/question/8828503
2. Learn more about calorimetry https://brainly.com/question/8168263
3. Learn more about heat of reaction https://brainly.com/question/10122365
Keywords: heat capacity, specific heat, calorimetry
