A snow maker at a resort pumps 260 kg of lake water per minute and sprays it into the air above a ski run. The water droplets freeze in the air and fall to the ground, forming a layer of snow. If all of the water pumped into the air turns to snow, and the snow cools to the ambient air temperature of -6.6°C, how much heat does the snow-making process release each minute? Assume the temperature of the lake water is 14.1°C, and use 2.00x103 J/(kg·C°) for the specific heat capacity of snow.
To calculate the heat released during the snow-making process, we need to consider the following steps:
1. Determine the amount of heat transferred from the lake water to lower its temperature to -6.6°C.
2. Calculate the heat released as the water freezes into snow.
Step 1: Calculating the heat transferred from the lake water to lower its temperature:
To lower the temperature of the lake water from 14.1°C to -6.6°C, we need to calculate the heat transferred using the specific heat capacity formula:
Q = m * c * ΔT
Where:
Q is the heat transferred
m is the mass of the water
c is the specific heat capacity of water (assumed to be 4.18x10^3 J/(kg·°C))
ΔT is the change in temperature
Given:
m = 260 kg
c = 4.18x10^3 J/(kg·°C)
ΔT = -6.6°C - 14.1°C = -20.7°C
Q1 = 260 kg * 4.18x10^3 J/(kg·°C) * (-20.7°C)
= -2.13x10^7 J
Note: The negative sign indicates heat has been transferred from the water.
Step 2: Calculating the heat released as the water freezes into snow:
When water freezes, it releases heat equal to the latent heat of fusion. For water, the specific latent heat of fusion is 3.34x10^5 J/kg.
The mass of the water that freezes into snow is equal to the mass of the water pumped into the air.
Given:
m = 260 kg
Latent heat of fusion, L = 3.34x10^5 J/kg
Q2 = m * L
= 260 kg * 3.34x10^5 J/kg
= 8.68x10^7 J
So, the heat released during the snow-making process each minute is:
Q_total = Q1 + Q2
= -2.13x10^7 J + 8.68x10^7 J
= 6.55x10^7 J
Therefore, the snow-making process releases approximately 6.55x10^7 J of heat each minute.