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

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

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

a) the particles of the chair are strongly held together.