Avg. temperature rise & energy conversion

[timothy997]

The problem is: The sun is 1.5x10^11 meters from the earth. Energy from the sun is received at the Earth's surface at the rate of 1.4 kilowatts per square meter. I'm trying to find two answers:
1. This energy flux from the Sun falls on a pond of water 100 square meters in area and 0.1 meter in depth. Assume all of this energy heats the water. Find the average temp. rise of the pond after 10^3 seconds
2. Determine the rate in kilograms per second at which the sun's mass is being converted to energy.




For 1, energy flux is Iota=power/area, power=work/time,
For 2, the SA of a sphere is 4pir^2, E=mc^2


So far I have p/1000 for energy flux and power=w/1000. I'm stuck on what to do next, but since the question gives you the distance between the Earth and sun, would w=fd help you find the solution? This question seems unusual compared to ones I've done in the past.

 

[physicshelppppp]

definition of a kilowatt

Remember that a kilowatt is 1000 watts, which is 1000 joules/second.

Hence, the flux is (1400 joules/second)/100 meters squared, which is 140 joules per meter squared second.

I am not so sure about the rest of the problem, but perhaps this information will help you. I will keep looking at it, though.

 

[ogb p]

I would approach this problem by first converting the given values into SI units. The distance between the Earth and sun, 1.5x10^11 meters, can be converted to kilometers by dividing by 1000, giving a value of 1.5x10^8 km. The energy received from the sun, 1.4 kilowatts per square meter, can be converted to watts by multiplying by 1000, giving a value of 1.4x10^3 watts per square meter.

1. To find the average temperature rise of the pond after 10^3 seconds, we can use the formula Q = mcΔT, where Q is the amount of heat transferred, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. We are given the area of the pond, 100 square meters, and the depth, 0.1 meters, which gives a volume of 10 cubic meters. Using the density of water, 1000 kg/m^3, we can calculate the mass of the water to be 10,000 kg. The specific heat capacity of water is 4.186 joules per gram per degree Celsius. Plugging in these values, we get:

Q = (10,000 kg)(4.186 J/g°C)(ΔT)

The power received from the sun is equal to the rate at which heat is transferred, so we can also write:

Q = (1.4x10^3 watts/m^2)(100 m^2)(10^3 seconds)

Setting these two equations equal to each other and solving for ΔT, we get:

ΔT = (1.4x10^3 watts/m^2)(100 m^2)(10^3 seconds) / (10,000 kg)(4.186 J/g°C)

ΔT = 33.5°C

Therefore, the average temperature rise of the pond after 10^3 seconds is 33.5 degrees Celsius.

2. To determine the rate at which the sun's mass is being converted to energy, we can use the famous equation E=mc^2, where E is the energy converted, m is the mass of the sun, and c is the speed of light. We know that the sun's mass is constantly being converted to energy through nuclear fusion, so we can calculate