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How much thermal energy (in J) is required to boil 2.65 kg of water at 100.0°C into steam at 145.0°C? The latent heat of vaporization of water is 2.26 ✕ 106 J/kg and the specific heat of steam is
2010
J
kg · °C
.
HINT
J

Respuesta :

Answer:

275494000 J

Explanation:

m = 2.65 kg

T1 = 100 C

T2 = 145 C

(Lv) Specific heat of steam = 2.26 x 10^6 J/Kg

c = 2010 J/Kg x C

Q1 = mLv = (2.65 kg)(2.26x10^-6)

                = 5989000 J

Heat required to raise temperature to T = 145 C is

Q2 = mc(delta T) = (2.65kg)(2.26 x 10^6 J/Kg)(145C - 100C)

                              = 269505000 J

Total thermal energy required:

Q = Q1 + Q2

   = 5989000 J + 269505000 J

   = 275494000 J

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