J.R. S. answered 10/11/20
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Ph.D. University Professor with 10+ years Tutoring Experience
C'mon, at least give it a try.
q = mC∆T
1201 J = (m)(0.45 J/gº)(97.5º)
Solve for m
Yesenia V.
asked 10/11/20If 1201 J of heat is available, what is the mass in grams of iron( specific heat = 0.45 J/g C) the can be heated from 22.5 Celsius to 120.0 Celsius
J.R. S. answered 10/11/20
Ph.D. University Professor with 10+ years Tutoring Experience
C'mon, at least give it a try.
q = mC∆T
1201 J = (m)(0.45 J/gº)(97.5º)
Solve for m
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