Question

In some countries, liquid nitrogen is used on dairy trucks instead of mechanical refrigerators. A (3.00-hour) delivery trip requires (200 L) of liquid nitrogen, which has a density of (808 kg/m^3). (a) Calculate the heat transfer necessary to evaporate this amount of liquid nitrogen and raise its temperature to (3.00°C). (Use (c_P) and assume it is constant over the temperature range.) This value is the amount of cooling the liquid nitrogen supplies. (b) What is this heat transfer rate in kilowatt-hours? (c) Compare the amount of cooling obtained from melting an identical mass of (0-°C) ice with that from evaporating the liquid nitrogen.

a) Determine the heat transfer necessary to evaporate liquid nitrogen.
b) Calculate the heat transfer rate in kilowatt-hours.
c) Compare the cooling effects of liquid nitrogen and melting ice.

Answer 1

This response explains how to calculate the heat transfer necessary to evaporate the liquid nitrogen, the heat transfer rate in kilowatt-hours, and compares the cooling effects of liquid nitrogen and melting ice.

Step-by-step explanation:

To calculate the heat transfer necessary to evaporate the liquid nitrogen, we need to find the energy required for (a) raising its temperature to 3.00 °C and (b) evaporating it.

For (a), we can use the equation Q = mcΔT, where Q is the heat transfer, m is the mass, c is the specific heat capacity, and ΔT is the temperature change. Since the heat transfer is to increase the temperature of the liquid nitrogen, the change in temperature is ΔT = (3.00 °C - (-196.00 °C)).

For (b), we can use the equation Q = mL, where Q is the heat transfer, m is the mass, and L is the latent heat of vaporization for liquid nitrogen, which is 199 kJ/kg.

The heat transfer rate in kilowatt-hours can be calculated using the formula Energy (kWh) = Q (J) / (3.6 * 10^6).

(c) We can compare the cooling effects by considering the energy required to melt an identical mass of 0 °C ice, which can be calculated using the formula Q = mL, where m is the mass and L is the latent heat of fusion for ice, which is 334 kJ/kg.