rate of absorption...

  • Imagine you are sipping tea or coffee while discussing various issues with a broad and diverse network of students, colleagues, and friends brought together by the common bond of physics, graduate school, and the physics GRE.

Posts: 3
Joined: Thu Dec 11, 2008 7:26 am

rate of absorption...

Postby dg2008 » Thu Dec 11, 2008 7:38 am

but how is rate of absorbtion worked out?

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 -> ???

User avatar
Posts: 1531
Joined: Thu Apr 13, 2006 2:47 pm

Re: rate of absorption...

Postby twistor » Thu Dec 11, 2008 9:40 am

You are given the rate of energy absorption per unit area, in watts. Recall that 1 W = 1 J/s. You know how much energy it takes to melt the ice and then raise the temperature, and you know the rate at which energy is being added to the ice/water. Use this information to find out how long it takes to add all of the required energy.

Posts: 3
Joined: Thu Dec 11, 2008 7:26 am

Re: rate of absorption...

Postby dg2008 » Thu Dec 11, 2008 9:46 am

so 49% of s'olar constent x area ->> 1370x3.9*10-2 = 53.43 ' or 49% of 1370?

and 194700 J + 11151 J = 205921 J ->> so is it this / by above?

608/7x0.039 = 26.1807

205921/26.1807=7865.37 s -> 131.09 minutes -> 2.31.09hrs

Return to “Physics Lounge”

Who is online

Users browsing this forum: No registered users and 1 guest