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Calculate the specific heat of the substance if 373 J is required to raise the temperature of a 312 gram sample by 25°C?

Sagot :

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)

mahima6824soni

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

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