Question

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?

Answer 1

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