Pls helpp An unknown substance with a mass of 23.5 grams absorbs 419.7 J while undergoing a temperature increase of 76.0°C. What is the specific heat of the substance?

The specific heat of the substance  : c = 0.235 J/g  °C

Given

Heat absorbed by substance = 419.7 J

m = mass = 23.5 g

Δt = Temperature difference : 76 °C

Required

The specific heat

Solution

Heat can be formulated

Q = m.c.Δt

Input the value :

419.7 = 23.5 x c x 76

c = 419.7 : (23.5 x 76)

c = 0.235 J/g  °C

the answer is 75.0%.