Question

15. A 1.00-kg block of copper at 20.0°C is dropped into a large vessel of liquid nitrogen at 77.3 K. How many kilograms of nitrogen boil away by the time the copper reaches 77.3K? (The specific heat of copper is 0.0920cal/g°C. The latent heat of vaporization of nitrogen is 48.0cal/g.)

Answer 1

First, we need to calculate the heat lost by the copper block as it cools from 20.0°C to 77.3K.

Q = mcΔT
Q = (1.00 kg)(0.0920 cal/g°C)(20.0°C - (-153.2°C))
Q = 1.00 kg * 0.0920 cal/g°C * 173.2°C
Q = 16.0088 cal

Next, we need to calculate the amount of nitrogen that will boil away as a result of the heat lost by the copper block.

Q = mL
16.0088 cal = (m)(48.0 cal/g)
m = 0.33435185 g

Finally, we convert the mass of nitrogen from grams to kilograms.

m = 0.33435185 g * 1 kg / 1000 g
m = 0.0003344 kg

Therefore, approximately 0.0003344 kg of nitrogen will boil away by the time the copper block reaches 77.3K.