Question

If it takes 22974 J to heat 96.5 g of a substance from 13.46°C to 73.86 °C, what is the specific heat of the substance?​

Answer 1

The specific heat capacity, C, of the substance is 3.94 $\frac{J}{g-^oC}$. 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

$q \ = \ mC(T_{f}-T_{i})$

where

q = amount of heat absorbed (usually in J)

m = amount of substance (in g)

C = specific heat capacity (in $\frac{J}{g-^oC}$)

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:

$C \ = \frac{q}{m(T_{f} \ - \ T_{i})}$

Plug in the values to the equation above to get the value of C.

$C \ = \ \frac{22974 \ J}{96.5 \ g \ (73.86 \ \ -13.46 \°C)} \\C \ = \ \frac{22974 \ J}{96.5 \ g \ (60.4\°C)} \\C \ = \ \frac{22974 \ J}{5828.6 \ g -\°C} \\C \ = \ 3.941598 \ \frac{J}{g -\°C}$

Since the least number of significant figures in the given is 3, then the final answer must have 3 significant figures, too. Therefore,

$\boxed {C = 3.94 \ \frac{J}{ g -\°C}}$

Learn More:

1. Learn more about heat capacity brainly.com/question/8828503

2. Learn more about calorimetry brainly.com/question/8168263

3. Learn more about heat of reaction brainly.com/question/10122365

Keywords: heat capacity, specific heat, calorimetry