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Q2=332*10=3320 kJ
Qc=210+3320=3530 kJ
m = 10 kg
ΔT = 0 - (-10)
ΔT = 10
1) ogrzanie lodu o 10 stopni
Q1 = m * Cw(cieplo wlasciwe lodu) * ΔT
Cw = 2100 [J-kgJ]
Q1 = 10 * 2100 * 10
Q1 =210 000 J
2) stopienie lodu
Ct =334 000 J kg-1
Q2 = Ct * m
Q2 = 334 000 * 10
Q2= 3 340 000 J
Q = Q1 + Q2
Q = 3 340 000 + 210 000
Q = 3 550 000 J