Question

1. How much heat is required to change 49.0 grams of ethanol ice at -114 ^oC to ethanol vapor at 78 ^oC? Ethanol has a heat capacity of 2.46.

2. A radiator has an emissivity of 0.825 and its exposed area is 9.375 m² The temperature of the radiator is 112.586 ^oC and the surrounding temperature is 4.718 ^oC. What is the heat flow rate from the radiator?

Answer 1

1. Now there will be three different heat absorption by ethanol ice to turn into ethanol vapor. First will be latent heat of fusion. Other will be heat absorbed by ethanol liquid to increase the temperature, and lastly latent heat of vaporization.

Now, Since ethanol melts at -114^o So the heat absorbed for fusion will be

Q₁ = mL_f = 49g x 108kJ/kg = 5292 J

After this the liquid ethanol will rise to temperature 78^o at which it will vaporize so,

Q₂ = mCdT = 49g x 2.46 J/gC^o x (78 -(-114)) = 23143.68 J

Now that it has reached 78^o and thus it will vaoprize by absorbing heat

Q₃ = mL_v = 49g x 855J/g = 41895J

Thus net heat absorbed by the ethanol ice to turn to vapor ethanol is

Q = Q₁ + Q₂ + Q₃ = 70330.68J = 70.33068 kJ

2) Heat flow rate of a radiation is given by

P = eA(T⁴ - T_s⁴) = (5.67 x 10^-8 J/s⋅m²⋅K⁴) x 0.825 x 9.375m² x [(112.586 + 273.15 K)⁴ - (4.718 + 273.15 K)⁴]

=> P = 7094.5343 J/s