Given:
m = 33.3 g
C = 4.184 J/g.oC
Ti = 0 oC
Tf = 50 oC
use:
Q = m*C*(Tf-Ti)
Q = 33.3*4.184*(50.0-0.0)
Q = 6966 J
Q = 6.97 KJ
Answer: 6.97 KJ
Calculate the amount of energy used when 33.3 grams of water at 0.00 °C is heated...
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