Question

A 34.87 g sample of a substance is initially at 27.1 C. after absorbing 1071 J of heat, the temperature of the substance is 145.0 C. what is the specific heat (SH) of the substance?

SH=___ J / g * C

Answer 1

Answer: .1044 J/g*C°

Explanation: The equation you need to use for this problem is :

c= q/m* ΔT

We are given

T1= 27.1 C°

T2= 145 C°

M= 34.87g

1071 J absorbed heatt

So let's solve specific heat

c= 1071J/ (87grams)*(145 C°- 27.1 C°)

c= 1071J/10,257.3 g*C°

c= .1044 J/g*C°

c=specific heat

Answer 2

`H= (1071)/(34.87)*145\\H= (155295)/(34.87) \\H= (15529500)/(3487) \\H=4453.542`