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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/KgThe question was posed to me in examination.The question is from Thermodynamics in division Thermodynamics of Chemistry – Class 11

Answer»

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.



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