Question

The total heat energy utilized for increasing the temperature by 4 degrees Kelvin in a 3 kgs substance is 100 KJ what is the specific heat capacity of that substance?

(a) 8.34 KJ/g-k

(b) 8.34 KJ/Kg-k

(c) 8.34 KJKg-k

(d) 8.34 KJ/Kg

The question was posed to me in examination.

The question is from Thermodynamics in division Thermodynamics of Chemistry – Class 11

Answer 1

Right answer is (b) 8.34 KJ/Kg-k

The best I can explain: The formula of heat energy is given by the expression: Q = mcΔT, m is the mass of the substance, c is the specific heat capacity and ΔT is the temperature difference. c = Q/mΔT; c = 100KJ/3kg(4k) = 8.34 KJ/Kg-k.