Answer :
Answer:
The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.
Explanation:
From Heat Transfer we determine that heat transfer rate due to electromagnetic radiation ([tex]\dot Q[/tex]), measured in BTU per hour, is represented by this formula:
[tex]\dot Q = \epsilon\cdot A\cdot \sigma \cdot (T_{s}^{4}-T_{b}^{4})[/tex] (1)
Where:
[tex]\epsilon[/tex] - Emissivity, dimensionless.
[tex]A[/tex] - Surface area of the student, measured in square feet.
[tex]\sigma[/tex] - Stefan-Boltzmann constant, measured in BTU per hour-square feet-quartic Rankine.
[tex]T_{s}[/tex] - Temperature of the student, measured in Rankine.
[tex]T_{b}[/tex] - Temperature of the bus, measured in Rankine.
If we know that [tex]\epsilon = 0.90[/tex], [tex]A = 16.188\,ft^{2}[/tex], [tex]\sigma = 1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}}[/tex], [tex]T_{s} = 554.07\,R[/tex] and [tex]T_{b} = 527.67\,R[/tex], then the heat transfer rate due to electromagnetic radiation is:
[tex]\dot Q = (0.90)\cdot (16.188\,ft^{2})\cdot \left(1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}} \right)\cdot [(554.07\,R)^{4}-(527.67\,R)^{4}][/tex]
[tex]\dot Q = 417.492\,\frac{BTU}{h}[/tex]
Under the consideration of steady heat transfer we find that the net energy transfer from the student's body during the 20 min-ride to school is:
[tex]Q = \dot Q \cdot \Delta t[/tex] (2)
Where [tex]\Delta t[/tex] is the heat transfer time, measured in hours.
If we know that [tex]\dot Q = 417.492\,\frac{BTU}{h}[/tex] and [tex]\Delta t = \frac{1}{3}\,h[/tex], then the net energy transfer is:
[tex]Q = \left(417.492\,\frac{BTU}{h} \right)\cdot \left(\frac{1}{3}\,h \right)[/tex]
[tex]Q = 139.164\,BTU[/tex]
The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.