Calculating Specific Heat: 50g of Unknown Substance at 25°C to 89.7°C

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To calculate the specific heat of a 50g unknown substance heated from 25°C to 89.7°C after absorbing 2.578 kJ of energy, the formula q = mcΔT is used, where q is the heat energy in Joules, m is the mass, c is the specific heat, and ΔT is the temperature change. The correct temperature change (ΔT) is 64.7°C. The confusion arose from mixing units, as q should be converted to Joules (2.578 kJ = 2578 J) for accurate calculations. The specific heat can be calculated by rearranging the formula to c = q / (mΔT). The discussion highlights the importance of unit consistency in thermodynamic calculations.
Jurrasic
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50 grams of a substance absorbed 2.578 kj of energy as it changed from 25 degrees celcius to 89.7
What is the specific heat of the unknown substance in J/gC ?

Is it correct to use :
q = m c delta T

q = specific heat
m = mass
c = ?
delta T here is 64.7

The answer seems really big? Is it :
(50grams) x (2578kj) x (64.7 C ) = answer ?
That seems wrong? Is there a more correct way to do it?
 
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You've got your quantities confused. q = heat put into or taken away from the substance in Joules. c = specific heat of the substance in J/gC.
 
SteamKing said:
You've got your quantities confused. q = heat put into or taken away from the substance in Joules. c = specific heat of the substance in J/gC.

LOL oh thanks :)~

Got it working now haha
 
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