A 3.00 kg block of copper at18.0°C is dropped into a large vesselof liquid nitro

Question

A 3.00 kg block of copper at18.0°C is dropped into a large vesselof liquid nitrogen at 77.3 K. How many kilograms of nitrogen boilaway by the time the copper reaches 77.3 K? (The specific heat ofcopper is 0.0920 cal/g·°C. The latent heat ofvaporization of nitrogen is 48.0 cal/g.)

A 3.00 kg block of copper at18.0°C is dropped into a large vesselof liquid nitrogen at 77.3 K. How many kilograms of nitrogen boilaway by the time the copper reaches 77.3 K? (The specific heat ofcopper is 0.0920 cal/g·°C. The latent heat ofvaporization of nitrogen is 48.0 cal/g.)

Explanation / Answer

the heat energy when block of copper reaches 77.3 K is Q = m * s * t ------------(1) m = 3.00 kg s = (0.0920 * 4.2/10^-3) J/kg.oC t = (t1 - t2) t1 = 18.0 oC = (18.0 + 273) K = 291 K t2 = 77.3 K the heat released when liquid nitrogen boils away Q1 = m1 * Lv -----------(2) m1 is mass of liquid nitrogen Lv is the latent heat of vaporization of nitrogenand has a value Lv= 48.0 cal/g = 48.0 * (4.2/10^-3)J/kg or Lv= 201.6 * 10^3 J/kg from (1) and (2) Q = Q1 m * s * t = m1 * Lv or m1 = (m * s * t/Lv) m1 is mass of liquid nitrogen Lv is the latent heat of vaporization of nitrogenand has a value Lv= 48.0 cal/g = 48.0 * (4.2/10^-3)J/kg or Lv= 201.6 * 10^3 J/kg from (1) and (2) Q = Q1 m * s * t = m1 * Lv or m1 = (m * s * t/Lv)

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