In a science lab, a student heats up a chemical from 10 °C to 25 °C which requires thermal energy of 30000 J. If mass of the object is 40 g, the specific heat capacity of the chemical would be

Group of answer choices

25 J /g* °C

75 J /g* °C

100 J /g* °C

50 J /g* °C

The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)

Explanation:

Step 1: Data given

Initial temperature = 10.0 °C

Final temperature = 25.0 °C

Energy required = 30000 J

Mass of the object = 40.0 grams

Step 2: Calculate the specific heat capacity of the object

Q = m* c * ΔT

⇒With Q = the heat required = 30000 J

⇒with m = the mass of the object = 40.0 grams

⇒with c = the specific heat capacity of the object = TO BE DETERMINED

⇒with ΔT = The change in temperature = T2 - T2 = 25.0 °C - 10.0°C = 15.0 °C

30000 J = 40.0 g * c * 15.0 °C

c = 30000 J / (40.0 g * 15.0 °C)

c = 50 J/g°C

The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)