O ile ciepła ogrzeje się 2 kg ołowiu, któremu dostarczono 1300 J ciepła? Ciepło właściwe ołowiu cPb=128 J/kg oC
m = 2kg
Q = 1300J
c = 128J/kg°C
Q = cmΔT => ΔT = Q/cm
ΔT = 1300J/2kg*128J/kg°C = 5,08°C
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Litterarum radices amarae sunt, fructus iucundiores
Pozdrawiam :)
Dane:Cw Ołowiu=128J/kg*oC
Q=1300J=1,3kJ
m=2kg===========================
Szukane:T=?
===========================
Wzór:Cw=Q/m*T
m*T=Q/Cw|:m
T=Q/Cw*m
===============================
T=1300J/128J/kg*oC*2kg
T=1300/256
T=5,078125oC
Odp:Ogrzeje się o 5,078125oC.Pozdrawiam kari9812
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m = 2kg
Q = 1300J
c = 128J/kg°C
Q = cmΔT => ΔT = Q/cm
ΔT = 1300J/2kg*128J/kg°C = 5,08°C
----------------------------------------------------------------------------------------------------
Litterarum radices amarae sunt, fructus iucundiores
Pozdrawiam :)
Dane:
Cw Ołowiu=128J/kg*oC
Q=1300J=1,3kJ
m=2kg
===========================
Szukane:
T=?
===========================
Wzór:
Cw=Q/m*T
m*T=Q/Cw|:m
T=Q/Cw*m
===============================
T=1300J/128J/kg*oC*2kg
T=1300/256
T=5,078125oC
Odp:Ogrzeje się o 5,078125oC.
Pozdrawiam kari9812