1.

Calculate the amount of energy needed to raise the temperature of 3.0 kg of steel from 20 °C to 120 °C. Thespecific heat capacity of steel is 490 J kg-1 K-1. please answer with a step by step with your answer and with all formulas filled in.​

Answer»

ss(M) = 3kgchange in TEMPERATURE(ΔT)= 120K-20K                                      = 100*C                                    c= 490 J Kg-1 K-1                                   Q= McΔT                                  490 × 3 × 100                                  147000 J Ans.Hence 14700 J of HEAT energy needed to raise the temperature of 3 kg of steel from 20K to 120K (or in degree celsius one and the same THING)



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