6. Following are actual data collected at the meteorological tower on Niwot Ridg

Question

6. Following are actual data collected at the meteorological tower on Niwot Ridge. Calculate the average loss of water from the snowpack in mm/day for the months of November (no liquid water in snowpack) and May (liquid water in snowpack). To answer this question, remember that watts (W) are equal to Joules per second (I/s) and that there are 1000 kilograms of water per cubic meter of water (10 points). o November, daily average latent heat flux was -77.11 W/m2. o May, daily average latent heat flux was-51.6 W/m2. o Take a guess at why the evaporative loss from the snowpack was greater in November than in May

Explanation / Answer

1 W = 1 J s-1

1 W m-2 = 0.0864 MJ m-2 day-1

2.45 MJ m-2 day-1 = 1 mm day-1

1 MJ m-2 day-1 = 1/2.45 mm day-1 = 0.408 mm day-1

For November:

77.11 W/m2 = 6.662 MJ m-2 day-1

1 MJ m-2 day-1 = 0.408 mm day-1

6.662 MJ m-2 day-1 = 2.71 mm day-1

b) For May 51.6 W/m2 = 4.4582 MJ m-2 day-1

1 MJ m-2 day-1 = 0.408 mm day-1

4.4582 MJ m-2 day-1 = 1.8189 mm day-1

c) Evaporative loss from the snowpack was greater in november than in May because  the saturation vapor pressure over and ice surface is slightly lower than over a water surface.

1 W = 1 J s-1

1 W m-2 = 0.0864 MJ m-2 day-1

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