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Chemistry 23 Online
OpenStudy (anonymous):

A sample of an unknown substance has a mass of 89.5 g. If 345.2 J of heat are required to heat the substance from 285 K to 305 K, what is the specific heat of the substance? Answer: ____ J/g•K

OpenStudy (abmon98):

E = m × c × θ E is the energy transferred in joules, J m is the mass of the substances in kg c is the specific heat capacity in J / kg °C θ (‘theta’) is the temperature change in degrees Celsius, °C 345.2=89.5*c*((305-273)-(285-273)) we subtract by -273 to change from kelvin to degree Celsius

OpenStudy (abmon98):

oh sorry i did a mistake dont change the unit leave it as it is because you need the units in kelvin iam sorry (305-285)

OpenStudy (anonymous):

Question. Is the equation you gave me "E = m × c × θ", the same as the equation "q=mCp△T"?

OpenStudy (aaronq):

it is, he just used different symbols

OpenStudy (somy):

true true

OpenStudy (anonymous):

Ah okay. I see now. Thank you.

OpenStudy (abmon98):

Your Welcome :)

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