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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

Sagot :

Since3122015idn
[tex]H= \frac{1071}{34.87}*145\\H= \frac{155295}{34.87} \\H= \frac{15529500}{3487} \\H=4453.542[/tex]

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

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