Datos:
Q=?
m=0,8kg
calor especifico Ce (agua)=4.186 J/kg.°c
T2=15 °c
T1= 5°c
t=600 seg
Solución:
Q=m.Ce.ΔT
Q=(0,8kg).(4.186 J/kg.°c).(15 °c - 5°c)
Q=33,5 kJ
P = Q/t
P=33,5 kJ/600 seg
P= 56 W
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Datos:
Q=?
m=0,8kg
calor especifico Ce (agua)=4.186 J/kg.°c
T2=15 °c
T1= 5°c
t=600 seg
Solución:
Q=m.Ce.ΔT
Q=(0,8kg).(4.186 J/kg.°c).(15 °c - 5°c)
Q=33,5 kJ
P = Q/t
P=33,5 kJ/600 seg
P= 56 W