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

Question

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

alekos · Answer

The formula for specific heat is  Q = cm x Temperature change

Where Q is the energy, m is the mass and c is the specific heat

just re-arrange to make c the subject of the formula and substitute the values given.

this will give you your answer

anonymous · Answer

c= -q x temperature change/m
c = 2165J x 35kelvin / 28.2
c = 2687
c = 2.69 x 10 ^3
?

alekos · Answer

No
c = Q/(m x Temp change)

alekos · Answer

you need to review your algebraic technique. what year are you in?