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How much heat, in calories, is needed to raise a 3.4g sample of iron 25ºC? (Specific Heat of Fe=.11 cal/gºC)?
In the equation q=mC∆T, I know that I must rearrange two variables. I also believe that 3.4g=m; .11cal/gºC=q; and 25ºC=q, but I am partially unsure. Would the equation set up be C=25/(3.4 x .11)? The answer I am getting is 66.8449, but I believe that change in heat is usually a large number. What step am I not getting correct?
Thank you for pointing out that the equation was already set up correctly. From the notes that I'm reading, I'm confusing q with the specific heat capacity.
2 Answers
- Anonymous7 years agoFavorite Answer
you are over complicating this
the specific heat tells you that .11 calories heats up one gram one degree C
so if you have 3.4 grams, multiply that by .11 to get calories needed to heat up 3.4 grams one degree, then multiply that by 25 to get calories needed to heat 3.4 grams 25 degrees
.11x3.4x25=9.35 calories, it seems small but iron being a metal and with a low specific heat, heats easily
- 7 years ago
just multiply everything…..you have everything you need. m=3.4g, C= .11 cal/g* ºC and ∆T= 25ºC that should give you your q which is heat. remember heat and temperature are not the same thing.
