A 1.0-kg block of copperat 30°C is dropped into a large insulated calorimeter containing liquid nitrogen at 77.3 K. How many kilograms of nitrogen boil away by the time the copper block reaches 77.3 K?The latent heat of vaporization of nitrogen is 199 J/g, its boiling point is 77.3 K, and the specific heat of copper is 0.386 J/(g°C)

Mass of Nitrogen that boils away = 0.4381 kg

Explanation:

Heat lost by the Copper block = (heat gained by the calorimeter) + (heat gained by Nitrogen liquid)

Heat lost by the Copper block = mCΔT

m = 1.00 kg,

C = specific heat capacity of Copper = 0.386 J/g.°C = 386 J/kg.K,

ΔT = change in temperature for Copper = (30+273.15) - 77.3 = 225.86 K

Heat lost by the Copper block = 1(386)(225.85) = 87178.1 J

Heat gained by Calorimeter = 0 J (since it is stated that the calorimeter is well insulated)

Heat gained by Nitrogen = mL

m = mass of Nitrogen that boils away = ?

L = Latent heat of vaporization of nitrogen = 199 J/g = 199000 J/kg

(Heat lost by the Copper block) = (heat gained by Nitrogen liquid)

87178.1 = m × 199000

m = (87178.1/199000)

m = 0.4381 kg

an electric circuit

option number c is the right answer.