Given 28.2 grams of an unknown substance, if the substance absorbs 2165 joules o

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Given 28.2 grams of an unknown substance, if the substance absorbs 2165 joules of energy and the temperature increases by 35 Kelvin, what is the specific heat of the substance?

2.14 x 106 J/g·K
2.19 x 100 J/g·K
5.73 x 10-4 J/g·K
2.69 x 103 J/g·K
Oct 18th, 2017

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Jun 23rd, 2015

Specific heat: Q = mcdT. 

Q=heat to change temp of substance = 2165J

m=mass of heated substance= 28.2 g

c=specific heat capacity = ?

dT=temp change = 35K

c = 2165/ (28.2)(35)

c = 2.19 x 100 J/g·K



Jun 23rd, 2015

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