Asample of an unknown substance has a mass of 0.158 kg. if 2,510.0 j of heat is required to heat the substance from 32.0°c to 61.0°c, what is the specific heat of the substance?

Given:

Mass of the unknown substance, m = 0.158 kg = 158 g

Heat absorbed Q = 2510.0 J

Initial temperature T1 = 32 C

Final temperature T2 = 61 C

To determine:

The specific heat of the substance

Explanation:

The heat absorbed to raise the temperature from T1 to T2 is-

Q = mcΔT

where c = specific heat

c = Q/mΔT = Q/m(T2-T1)

c = 2510 J/158 g * (61-32) C = 0.548 J/gC

Ans: Specific of the substance is 0.548 J/gC