Answer:
Approximately 45.31 grams of nitrogen boil away by the time the bismuth reaches 80.0 K.
Step-by-step explanation:
To find out how many kilograms of nitrogen boil away, we need to calculate the amount of heat transferred from the bismuth to the nitrogen, which causes the nitrogen to boil.
The formula for calculating the heat transferred is:
Q = mcΔT
Where:
Q = Heat transferred (in calories)
m = Mass of the substance (in grams)
c = Specific heat of the substance (in cal/g °C)
ΔT = Change in temperature (in °C)
First, we'll calculate the heat transferred from the bismuth to reach the temperature of the liquid nitrogen:
m_bismuth = 1.50 kg = 1500 grams (since 1 kg = 1000 grams)
c_bismuth = 0.029 cal/g °C
ΔT_bismuth = (80.0 K - 30.0°C) = 50.0 °C
Q_bismuth = m_bismuth * c_bismuth * ΔT_bismuth
Q_bismuth = 1500 g * 0.029 cal/g °C * 50.0°C
Q_bismuth = 2175 cal
Now, we need to calculate the heat required to vaporize the nitrogen at its boiling point:
latent_heat_nitrogen = 48 cal/g
Q_nitrogen = latent_heat_nitrogen * m_nitrogen
Since the nitrogen starts at 80.0 K and ends at its boiling point (77.36 K at atmospheric pressure), the change in temperature is 77.36 K - 80.0 K = -2.64 K. However, we're interested in the absolute value of the change in temperature since the latent heat of vaporization is a positive value. So, ΔT_nitrogen = |(-2.64 K)| = 2.64 K.
Now, we need to find the mass of nitrogen that boils away, which we'll call m_boil_nitrogen:
Q_nitrogen = m_boil_nitrogen * latent_heat_nitrogen
m_boil_nitrogen = Q_nitrogen / latent_heat_nitrogen
m_boil_nitrogen = 2175 cal / 48 cal/g
m_boil_nitrogen ≈ 45.3125 g
So, approximately 45.31 grams of nitrogen boil away by the time the bismuth reaches 80.0 K.