64.2k views
4 votes
A 54.7 g sample of polystyrene is put into a calorimeter. Calculate the specific heat capacity of polystyrene according to this experiment. Be sure your answer is rounded to the correct number of significant digits.

1 Answer

6 votes

Final answer:

The specific heat capacity of the metal is calculated using the formula q = mcΔT, which yields a value of 0.453 J/g°C when solving with the given data. To identify the metal, this value should be compared to known specific heat values for various metals.

Step-by-step explanation:

To calculate the specific heat capacity of the metal, we will use the formula:

q = mcΔT

Where q is the heat absorbed or released (in joules or calories), m is the mass (in grams), c is the specific heat capacity (J/g°C or cal/g°C), and ΔT is the change in temperature (in degrees Celsius).

According to the data provided, the metal has a mass of 217 grams, absorbs 1.43 kJ of heat (which is equal to 1430 J), and experiences a temperature increase from 24.5°C to 39.1°C. Thus, the ΔT (change in temperature) is 39.1°C - 24.5°C = 14.6°C.

Now we can rearrange the specific heat capacity formula to solve for c:

c = q / (mΔT)

c = 1430 J / (217 g × 14.6°C)

After calculation:

c = 0.453 J/g°C

The specific heat capacity of the metal is therefore 0.453 J/g°C, rounded to three significant digits reflecting the precision of the given data.

To predict the identity of the metal, you can compare this calculated specific heat value to a table of specific heats for various metals.

User BueKoW
by
7.4k points