Answer :
Explanation:
Q = 373J
∆∅ = 25°C
m = 312g
c = x
Q = mc∆∅
373 = 312(x)(25)
373 = 7800x
x = 373/7800
x = 0.0478J/(g°C)
The specific heat of the substance if 373 J is required to raise teh temperature of a 312 gram sample by 25°C is 0.0478 J/g
What is specific heat?
The amount of energy needed to raise the temperature of one gram of a substance by one degree Celsius.
By the formula of specific heat
Q = mc∆∅
Q = 373J
m = 312g
∆∅ = 25° C
Putting the values in the equation
[tex]373 = 312\times c \times (25)\\\\373 = 7800c\\c= \dfrac{373}{7800} = 0.0478J/(g \circ C)[/tex]
Thus, the specific heat of the substance if 373 J is required to raise teh temperature of a 312 gram sample by 25°C is 0.0478 J/g
Learn more about specific heat
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