The sun radiates energy at the rate of 3.8x10^26 W from its 5500°C surface into the empty dark space (fraction of radiation to earth and other planets is negligible).The effective temperature of deep space is -270°C.
a. What is the increase in entropy in one day due to this heat transfer.?
b.how much work is made in available?
Thank you for answering my questions.
To calculate the increase in entropy and the work made available, we need to use the equation for entropy change and the equation for work done respectively.
a. Increase in entropy:
The equation for entropy change is given by:
ΔS = Q / T
Where:
ΔS is the change in entropy,
Q is the heat transferred, and
T is the temperature in Kelvin.
To calculate the increase in entropy in one day, we need to convert the temperatures to Kelvin and the energy radiated from power (Watts) to heat transferred (Joules) by multiplying by the time in seconds.
Given:
Solar energy radiated by the sun = 3.8x10^26 W
Time = 1 day = 24 hours = 24*60*60 seconds
Converting the temperature to Kelvin:
The surface temperature of the sun (5500°C) can be converted to Kelvin by using the formula K = °C + 273.15.
So, the temperature of the sun = 5500°C + 273.15 = 5773.15 K.
Plugging in the values:
Q = Power * Time = 3.8x10^26 * (24*60*60) Joules
T = Temperature of the sun = 5773.15 K
Now we can calculate the change in entropy:
ΔS = (3.8x10^26 * (24*60*60)) / 5773.15
b. Work made available:
The equation for work done is:
W = Q - ΔU
Where:
W is the work done,
Q is the heat transferred,
ΔU is the change in internal energy.
In this case, we assume no change in internal energy as the system (deep space) is at a constant temperature. Therefore, the change in internal energy (ΔU) is zero, and the equation simplifies to:
W = Q
Substituting the value of Q:
W = 3.8x10^26 * (24*60*60) Joules
Calculating the values will give you the specific values for the increase in entropy and work made available due to the heat transfer in one day.