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Sagot :
[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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