zapisz, ile ciepła odda 1kg pary wodnej o temp. 100 st. C, zamieniając się całkowicie w lód o temp. 0 st. C.
m=1kg
t₁=100⁰C
t₂=0⁰C
cp=2260000J/kg
cww=4200J/kg*^C
cwl=2100J/kg*^C
ct=335000J/kg
Q₁=m*cp
Q₁=1kg*2260000J/kg=2260000J
Q₂=m*cww*Δt
Q₂=1kg*4200J/kg*^C*100⁰C=420000J
Q₃=m*ct
Q₃=1kg*335000J/kg=335000J
Q=Q₁+Q₂+Q₃ = 2260000J+420000J+335000J=3 015 000J= 3015kJ
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m=1kg
t₁=100⁰C
t₂=0⁰C
cp=2260000J/kg
cww=4200J/kg*^C
cwl=2100J/kg*^C
ct=335000J/kg
Q₁=m*cp
Q₁=1kg*2260000J/kg=2260000J
Q₂=m*cww*Δt
Q₂=1kg*4200J/kg*^C*100⁰C=420000J
Q₃=m*ct
Q₃=1kg*335000J/kg=335000J
Q=Q₁+Q₂+Q₃ = 2260000J+420000J+335000J=3 015 000J= 3015kJ