Question

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

Answer 1

Heat = mC(t2 - t1)
So, 2510 j = 0.158kg
1000g/1kg * C * 61.0 - 32degrees C
therefore, C = 0.5478 j/g degrees C



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