# Algebraic Thermodynamics - Power absorbed and emitted by human body

1. Jul 28, 2011

### VinnyCee

1. The problem statement, all variables and given/known data

Problem 1

The emissivity of the human skin is 97.0 percent. Use 35.0 °C for the skin temperature and approximate the human body by a rectangular block with a height of 1.97 m, a width of 41.0 cm and a length of 34.5 cm.

a) Calculate the power emitted by the human body.

Fortunately our environment radiates too. The human body absorbs this radiation with an absorbance of 97.0 percent, so we don't lose our internal energy so quickly.

b) How much power do we absorb when we are in a room where the temperature is 26.5 °C?

c) How much energy does our body lose in one second?

Problem 2

n = 2.19 mol of Hydrogen gas is initially at T = 342 K temperature and pi = 1.68×105 Pa pressure. The gas is then reversibly and isothermally compressed until its pressure reaches pf = 9.14×105 Pa.

a) What is the volume of the gas at the end of the compression process?

b) How much work did the external force perform?

c) What would be the temperature of the gas, if the gas was allowed to adiabatically expand back to its original pressure?

2. Relevant equations

Problem 1

$$P\,=\,\epsilon\,\sigma\,A\,T^4$$

Problem 2

$$P\,V\,=\,n\,R\,T$$

$$C\,=\,\frac{p_f\,V_f}{T}$$

$$T\,=\,\frac{p_i\,V_i}{C}$$

3. The attempt at a solution

Problem 1

$$A\,=\,2\,a\,b\,+\,2\,b\,c\,+\,2\,a\,c$$

$$A\,=\,2\,\left(1.97\,m\right)\,\left(0.41\,m \right)\,+\,2\,\left(0.41\,m\right)\,\left(0.345\,m \right)\,+\,2\,\left(1.97\,m\right)\,\left(0.345\,m \right)$$

$$A\,=\,3.2576\,m^2$$

$$T_e\,=\,35\,+\,273.15\,=\,308.15\,K$$

$$T_a\,=\,26.5\,+\,273.15\,=\,299.65\,K$$

a)

$$P_e\,=\,\epsilon\,\sigma\,A\,T_e^4\,=\,\left(0.97 \right)\,\left(5.67\,X\,10^{-8}\,\frac{W}{m^2\,K^4}\right)\,\left(3.2576\,m^2 \right)\,\left(308.15\,K\right)^4$$

$$P_e\,=\,1615.5\,W$$

b)

$$P_a\,=\,\alpha\,\sigma\,A\,T_a^4\,=\,\left(0.97 \right)\,\left(5.67\,X\,10^{-8}\,\frac{W}{m^2\,K^4}\right)\,\left(3.2576\,m^2 \right)\,\left(299.65\,K\right)^4$$

$$P_a\,=\,1444.5\,W$$

c)

$$P\,=\,P_a\,-\,P_e\,=\,1444.5\,W\,-\,1615.5\,W\,=\,-171\,W$$

Therefore, a dude (possibly a http://www.msnbc.msn.com/id/43866502/ns/technology_and_science-science/" [Broken]), who is freezing his wey-bowls off ($95^{\circ}\,F$ body temperature!) loses...

$$171\,\frac{J}{s}$$

Problem 2

$$n\,R\,T\,=\,\left(2.19\,mol\right)\,\left(8.314\, \frac{J}{mol\,K}\right)\,\left(342\,K\right)\,=\,6227\,J$$

a)

$$p_f\,V_f\,=\,n\,R\,T$$

$$V_f\,=\,\frac{n\,R\,T}{p_f}\,=\,\frac{6227\,J}{1.68\,X\,10^5\,Pa}$$

$$V_f\,=\,0.5691\,m^3$$

b)

$$V_i\,=\,\frac{n\,R\,T}{p_i}\,=\,\frac{6227\,J}{1.68\,X\,10^5\,Pa}\,=\,0.1046\,m^3$$

$$W_{i\,-\,f}\,=\,n\,R\,T\,ln\,\frac{V_f}{V_i}\,=\,10548\,J$$

c)

$$C\,=\,\frac{p_f\,V_f}{T}\,=\,\frac{\left(9.14\,X\,10^5\,Pa\right)\,\left(0.5691\,m^3\right)}{\left(342\,K\right)}\,=\,1566$$

$$T\,=\,\frac{p_i\,V_i}{C}\,=\,\frac{\left(1.68\,X\,10^-5\,Pa\right)\,\left(0.1046\,m^3\right)}{1566}$$

$$T\,=\,11.22\,K$$

Which is really freezing!

Does this all seem correct? Or are there some things that I am missing?

Last edited by a moderator: May 5, 2017