Question

On a hot dry day, evaporation from a lake has just enough heat transfer to balance the 1.00 kW/m² of incoming heat from the Sun. What mass of water evaporates in 1.00 h from each square meter? Explicitly show how you follow the steps in the Problem-Solving Strategies for the Effects of Heat Transfer.

a) 1.0 kg
b) 2.0 kg
c) 3.0 kg
d) 4.0 kg

Answer 1

The mass of water evaporated in 1 hour from each square meter is 1.59 kg/m².

Step-by-step explanation:

To find the mass of water evaporated in 1.00 hour from each square meter, we need to use the Problem-Solving Strategies for the Effects of Heat Transfer. On a hot dry day, the incoming heat from the Sun is 1.00 kW/m². This incoming heat is balanced by the rate of evaporation.

We know that the energy needed for evaporation is Q = mL, where Q is the energy, m is the mass of water, and L is the latent heat of vaporization.

The incoming heat from the Sun can be converted to energy using the equation:

1.00 kW/m² = 1,000 J/s/m².

Since 1 watt is equivalent to 1 joule/second, we can conclude that the rate of evaporation is 1,000 J/s/m².

In 1 hour, there are 3,600 seconds.

So, the total energy available for evaporation in 1.00 hour is 1,000 J/s/m² × 3,600 seconds = 3,600,000 J/m².

Now, let's find the mass of water using the equation: Q = mL.

Rearranging the equation, we have: m = Q/L.

The latent heat of vaporization for water is approximately 2,260,000 J/kg.

So, the mass of water evaporated in 1.00 hour from each square meter is:

m = 3,600,000 J/m² / 2,260,000 J/kg = 1.59 kg/m².

Therefore, the answer is not given in the options provided.