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
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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