5. 54 J of heat is required to raise the temperature of a 2.47g unknown substance from 17.10C to 46.70C. a) Calculate the specific heat of the substance b) What is the substance

The specific heat of the substance is 0.739 J/g°C

The substance can be glass or SiO2

Explanation:

Step 1: Data given

Heat required = 54 J

Mass of the substance = 2.47 grams

Initial temperature = 17.10 °C

Final temperature = 46.70 °C

Step 2: Calculate the specific heat of the substance

Q = m*c¨*ΔT

⇒with Q = the heat required =54 J

⇒with m = the mass of the substance = 2.47 grams

⇒with c = the specific heat of the substance = TO BE DETERMINED

⇒with ΔT = the change of temperature = T2 - T1 = 46.70 - 17.10 = 29.60 °C

c = Q / (m*ΔT)

c = 54 / (2.47 * 29.60)

c = 0.739 J/g°C

The specific heat of the substance is 0.739 J/g°C

The substance can be glass or SiO2

l dont know l am s sorry

geosphere and hydrosphere.

this is because the geosphere has to do with geological events, while the hydrosphere controls all water on earth.