Calculate the experimental specific heat capacity of an object of mass 1.32 kg, given that the object releases 1.95 kJ of heat when its temperature decreases by 19.5°C. HELP GONNA FAIL THE SEMESTER HELPPO

0.076 J/gºC.

Explanation:

From the question given above, the following data were obtained:

Mass (M) of object = 1.32 kg

Heat (Q) released = –1.95 kJ

Change in temperature (ΔT) = –19.5°C (decrease in temperature)

Specific heat capacity (C) =?

Next, we shall convert 1.32 kg to grams (g). This can be obtained as follow:

1 kg = 1000 g

Therefore,

1.32 kg = 1.32 kg × 1000 g / 1 kg

1.32 kg = 1320 g

Next, we shall convert –1.95 kJ to joule (J). This can be obtained as follow:

1 kJ = 1000 J

Therefore,

–1.95 kJ = –1.95 kJ × 1000 / 1 kJ

–1.95 kJ = –1950 J

Finally, we shall determine the specific heat capacity of the object. This can be obtained as follow:

Mass (M) of object = 1320 g

Heat (Q) released = –1950 J

Change in temperature (ΔT) = –19.5°C (decrease in temperature)

Specific heat capacity (C) =?

Q = MCΔT

–1950 = 1320 × C × –19.5

–1950 = –25740 × C

Divide both side by –25740

C = –1950 / –25740

C = 0.076 J/gºC

Thus, the specific heat capacity of the object is 0.076 J/gºC.

if the concentration of products is decreased, the equilibrium will shift toward the products). add an inert gas (one that is not involved in the reaction) to the constant-volume reaction mixture: this will increase the total pressure of the system, but will have no effect on the equilibrium condition.

explanation:

the answer to your question is c