High School

37.9 grams of an unknown substance undergoes a temperature increase of 25.0°C after absorbing 989 J. What is the specific heat of the substance?

Answer :

The specific heat of the substance : c = 1.044 J/g °C
J/g °C


Further explanation

Given


Heat absorbed by substance = 989 J


m = mass = 37.9 g


Δt = Temperature difference : 25 °C


Required


The specific heat


Solution


Heat can be formulated


Q = m.c.Δt


Input the value :

989 = 37.9 x c x 25

c = 989 : (37.9 x 25)

c = 1.044 J/g °C


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