Question

An electric range burner weighing **646.0** grams
is turned off after reaching a temperature of
**466.2**°C, and is allowed to cool down to
**23.9**°C.

Calculate the specific heat of the burner if all the heat evolved
from the burner is used to heat **566.0** grams of
water from **23.9**°C to **82.5**°C.

Answer: ________ J/g°C

Answer #1

1st calculate the heat required to heat the water

we have:

m = 566 g

C = 4.184 J/g.oC

Ti = 23.9 oC

Tf = 82.5 oC

we have below equation to be used:

Q = m*C*(Tf-Ti)

Q = 566.0*4.184*(82.5-23.9)

Q = 138773 J

This heat must be supplied by burner

we have:

Q = -138773 J

m = 646 g

Ti = 466.2 oC

Tf = 23.9 oC

we have below equation to be used:

Q = m*C*(Tf-Ti)

-138773.0 = 646.0*C*(23.9-466.2)

C = 0.486 J/g.oC

Answer: 0.486 J/g.oC

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