Ask a New Question

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?
6 years ago

Answers

Answered by DrBob222
q = 4180 = mass H2O x specific heat H2O x (Tfinal0Tiitial)
specific heat H2O is 4.18 J/oC
6 years ago
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
3 years ago

Related Questions

Hello. Please help me with grammar. I know that "consider" is often used with "be" (for example, I... Hello. Please help me with grammar. 1)Which is better, one or a: "more than one third of Italy's po... 1 gram of C3H8 gas and 1 gram of O2 gas are delivered to a metal sphere with a fixed volume of 1 L.... how many gram of ice at 0 c can be melted by the heat released when one gram of steam condense to wa... 16 gram of CaC2 and 18 gram of H2O were mixed in the given reaction. CaC2 + 2H2O it yields C2H2 + C... 16 gram of CaC2 and 18 gram of H2O were mixed in the given reaction, CaC2 + 2H2O it gives C2H2 + Ca... 16 gram of CaC2 and 18 gram of H2O were mixed in the given reaction, CaC2 + 2H2O it gives C2H2 + Ca... 16 gram of CaC2 and 18 gram of H2O were mixed in the given reaction, CaC2 + 2H2O it gives C2H2 + Ca... Gram positive and gram negative What happens if you eat a gram of uranium?
Ask a New Question
Archives Contact Us Privacy Policy Terms of Use