Jaka była masa gliceryny , jeżeli po dostarczeniu jej energii cieplnej równej 200kJ jej temp. podniosła się z 0stop.cel do 120sto.cel ?
martu94
Q = 200 kJ = 200000 J T1 = 0 C T2 = 120 C delta T = 120 C cw = 2430 J/kg*C m = ? Q = m*cw*delta T m = Q / cw *delta T m = 200000 J/2430 J/kg *C* 120 C m = 200000 J /291600 J/kg m = 0,69 kg
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kodyr
Q = 200 kJ = 200000 J T1 = 0 C T2 = 120 C deltaT = 120 C cw = 2430 J/kg * C
m = ?
Q = m*cw*delta T m = Q / cw *delta T m = 200000 J/2430 J/kg *C* 120 C m = 200000 J /291600 J/kg m = 0,69 kg
T1 = 0 C
T2 = 120 C
delta T = 120 C
cw = 2430 J/kg*C
m = ?
Q = m*cw*delta T
m = Q / cw *delta T
m = 200000 J/2430 J/kg *C* 120 C
m = 200000 J /291600 J/kg
m = 0,69 kg
T1 = 0 C
T2 = 120 C
deltaT = 120 C
cw = 2430 J/kg * C
m = ?
Q = m*cw*delta T
m = Q / cw *delta T
m = 200000 J/2430 J/kg *C* 120 C
m = 200000 J /291600 J/kg
m = 0,69 kg