A 57.35 g sample of a substance is initially at 26.1 ∘C. After absorbing 2551 J of heat, the temperature of the substanc...
A 57.35 g sample of a substance is initially at 26.1 ∘C. After absorbing 2551 J of heat, the temperature of the substance is 159.0 ∘C. What is the specific heat (SH) of the substance?
SH=_____Jg⋅∘C
Homework Answers
Given:
Q = 2551 J
m = 57.35 g
Ti = 26.1 oC
Tf = 159 oC
use:
Q = m*C*(Tf-Ti)
2551.0 = 57.35*C*(159.0-26.1)
C = 0.335 J/g.oC
Answer: 0.335 J/g.oC
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