Question

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

Answer 1

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

The substance can be glass or SiO2

Step-by-step 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

Answer 2

C = 0.7385J/g°C

Step-by-step explanation:

Heat energy (Q) = 54J

Mass (m) = 2.47g

Initial temperature (T1) = 17.10°C

Final temperature (T2) = 46.70°C

Specific heat capacity of substance (c) = ?

Heat energy of a substance (Q) = Mc∇T

Q = heat energy

M = mass of the substance

C = specific heat capacity of the substance

∇T = change in temperature of substance = T2 - T1

Q = mc∇T

Q = mc (T2 - T1)

54 = 2.47 * c * (46.70 - 17.10)

54 = 2.47 * 29.6 * c

54 = 73.112c

C = 54 / 73.112

C = 0.7385

Specific heat capacity of the substance is 0.7385J/g°C

B.

The specific heat capacity of the above substance does not match any metal in the periodic table, however the closest to it was potassium with a value of 0.7536J/g°C and silicone with value of 0.711756J/g°C