Asked by Adora
If the temperature of 50.0 grams of water was raised to 50°c by the addition of 4180J of heat energy. What was the initial temperature of the water?
Answers
Answered by
DrBob222
q = 4180 = mass H2O x specific heat H2O x (Tfinal0Tiitial)
specific heat H2O is 4.18 J/oC
specific heat H2O is 4.18 J/oC
Answered by
I’m not giving you my name
Q = m • C • ΔT
Q = 4180J
m = 50.0g
C = 4.184 J/g*C
ΔT = (50*C - T(initial))
4180 = 50.0 • 4.184 • (50 - T)
4180 = 209.2 (50 - T)
4180 = -209.2T + 10460
-209.2T = -6280
T(initial) = 30.0*C
Q = 4180J
m = 50.0g
C = 4.184 J/g*C
ΔT = (50*C - T(initial))
4180 = 50.0 • 4.184 • (50 - T)
4180 = 209.2 (50 - T)
4180 = -209.2T + 10460
-209.2T = -6280
T(initial) = 30.0*C
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