Virtually all of the energy used on Earth derives originally from the Sun and solar radiation is intercepted at 1370 W m−2 at the top of the atmosphere, and that 49% of this radiation is absorbed by the Earth’s surface. Assuming that a shallow dish containing ice or water, placed on the Earth’s surface, absorbs the same fraction of solar radiation as the surface itself, at what rate is energy absorbed from the Sun by such a container of surface area 0.039 m2?
So far I have
solar constent x area ->> 1370x3.9*10-2 = 53.43
then I have
A student, keen to investigate the energy transformations, places a shallow dish of ice outside in the sun. The ice has a mass of 0.59 kg and a surface area of 0.039 m2. Assuming that the ice is originally at 0 °C and using your answer to part (a), for how long does the student have to wait until all the ice is melted and the temperature of the resulting water has reached 4.5 °C?
You may assume that the specific heat capacity of water, c, is 4.2 × 103 J kg−1 °C−1 and that the specific latent heat of melting of water, Lf, is 3.3 × 105 J kg−1 .
I came up with the following:
melting -> q=Lfm -> 0.59 x 3.3x105 = 194700 J
temp raise -> q=mcT -> 0.59 x 4.2x103 x 4.5 = 11151 J
length of time -> ???