Question

The tea will also cool as a result of evaporation. Suppose the evaporation rate from the tea vessel is 5×10

−4

kg/m

2

/s and occurs over the tea surface area of 80 cm

2
. How much more will the temperature of the tea change as a result of evaporative cooling over 10 minutes? Note that you can assume the mass change in tea due to evaporation is negligible here

Answer 1

Step-by-step explanation:

There is some information missing, such as the mass of the tea that is cooling due to evaporation, and its initial temperature. I've attached a worksheet with the key calculations, but the numbers assume both a starting mass of 500 grams and an initial temperature of 70 degrees C.

There are a lot of calculations, so maintaining units is important to insure correct steps.

Lines 1 - 8 collect what we know and begin the process of converting units, as necessary. We need both the heat of evaporation and the specific heat of water, as shown in lines 7 and 11. The evaporation rate is in m^2, but the value for the surface are is in cm^2, sio a conversion to mm^2 is necessary (Line 3). We also need seconds, not minutes (Line 5).

We can now calculate the kg of water that evaporates in 10 minutes (Line 6).

The heat of vaporization of water is given on Line 7. We will assume both 500g of tea and a starting temperature of