Ciepło właściwe stali cw≈450·J/kg⁰C.Ile ciepła trzeba dostarczyć, aby ogrzać 2kg stali o 20⁰C.
c = 450J/kg°C
m = 2kg
ΔT = 20°C
Q=?
Q = cmΔT
Q = 420J/kg°C x 2kg x 20°C = 16800J = 16,8kJ
W razie pytań pisz
I hope I helped
Pozdrawiam Marco12 :)
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c = 450J/kg°C
m = 2kg
ΔT = 20°C
Q=?
Q = cmΔT
Q = 420J/kg°C x 2kg x 20°C = 16800J = 16,8kJ
W razie pytań pisz
I hope I helped
Pozdrawiam Marco12 :)