What is the specific heat of a substance if a mass of 10.0 kg increases in temperature from 10.0°c to 70.0°c when 2,520 j of heat is applied?
0.00420 j/(gi°c)
0.00661 j/(gi°c)
238 j/(gi°c)
252 j/(gi°c)

We use the expression Heat = mC(Tf-Ti) where m is the mass of the sample, C is the specific heat and Tf, Ti are the final and initial temperature to calculate the problem above. We do as follows:

2520 = (10000)C ( 70-10)
C = 0.00420 J / g °C <FIRST OPTION 

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