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souacarol
@souacarol
October 2020
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questão de física sobre calorimetria (o gabarito se encontra a frente)
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GFerraz
M = 100 g
a)
No estado líquido:
ΔQ = 2000 cal
ΔT = 140-60 = 80ºC
c = Q/m.ΔT
c = 2000/100.80
c = 2/8
c = 0,25 cal/gºC
b)
ΔQ = 5200 - 2000 = 3.200 cal
Q = m.L
L = Q/m
L = 3200/100
L = 32 cal/g
c)
ΔQ = 6400 - 5200 = 1200 cal
ΔT = 60ºC
Q = m.c.ΔT
1200 = 100.60.c
1200 = 6000.c
c = 1200/6000
c = 12/60
c = 0,2 cal/gºC
3 votes
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a)
No estado líquido:
ΔQ = 2000 cal
ΔT = 140-60 = 80ºC
c = Q/m.ΔT
c = 2000/100.80
c = 2/8
c = 0,25 cal/gºC
b)
ΔQ = 5200 - 2000 = 3.200 cal
Q = m.L
L = Q/m
L = 3200/100
L = 32 cal/g
c)
ΔQ = 6400 - 5200 = 1200 cal
ΔT = 60ºC
Q = m.c.ΔT
1200 = 100.60.c
1200 = 6000.c
c = 1200/6000
c = 12/60
c = 0,2 cal/gºC