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Mikaellesan
@Mikaellesan
October 2020
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Um quilograma de glicerina,de calor específico 0,6 cal/g ºC,inicialmente a -30 C,recebe 12.000 cal de uma fonte.Determine a temperatura final da glicerina.
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pjfaria
Verified answer
1kg = 1000g
Q = m * c * /\T
12.000 = 1000 * 0,6 * (tf - (-30)
12000 = 600tf + 18.000
-6000 = 600tf
-6000/600 = tf
tf = -10°C
************
18 votes
Thanks 46
JimmyNeutron
Verified answer
1 kg corresponde a 1000 g
Q = mcΔt
12000 = 1000.0,6.(Tf-(-30))
12000 = 600 (Tf+30)
12000 = 600 Tf + 18000
-600 Tf = 18000 - 12000
-600 Tf = 6000
Tf = 6000 / -600
Tf = -10 ºC
14 votes
Thanks 36
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Lista de comentários
Verified answer
1kg = 1000gQ = m * c * /\T
12.000 = 1000 * 0,6 * (tf - (-30)
12000 = 600tf + 18.000
-6000 = 600tf
-6000/600 = tf
tf = -10°C
************
Verified answer
1 kg corresponde a 1000 gQ = mcΔt
12000 = 1000.0,6.(Tf-(-30))
12000 = 600 (Tf+30)
12000 = 600 Tf + 18000
-600 Tf = 18000 - 12000
-600 Tf = 6000
Tf = 6000 / -600
Tf = -10 ºC