Question

(35-39) A student travels on a school bus in the middle of winter from home to school. The school bus temperature is 68.0° F. The student's skin temperature is 94.4° F. Determine the net energy transfer from the student's body during the 20.00 min ride to school due to electromagnetic radiation. Note: Skin emissivity is 0.90, and the surface area of the student is 1.50m2.

Answer 1

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

Step-by-step explanation:

From Heat Transfer we determine that heat transfer rate due to electromagnetic radiation (
`\dot Q`), measured in BTU per hour, is represented by this formula:

`\dot Q = \epsilon\cdot A\cdot \sigma \cdot (T_(s)^(4)-T_(b)^(4))` (1)

Where:

`\epsilon` - Emissivity, dimensionless.

`A` - Surface area of the student, measured in square feet.

`\sigma` - Stefan-Boltzmann constant, measured in BTU per hour-square feet-quartic Rankine.

`T_(s)` - Temperature of the student, measured in Rankine.

`T_(b)` - Temperature of the bus, measured in Rankine.

If we know that
`\epsilon = 0.90`,
`A = 16.188\,ft^(2)`,
`\sigma = 1.714* 10^(-9)\,(BTU)/(h\cdot ft^(2)\cdot R^(4))`,
`T_(s) = 554.07\,R` and
`T_(b) = 527.67\,R`, then the heat transfer rate due to electromagnetic radiation is:

`\dot Q = (0.90)\cdot (16.188\,ft^(2))\cdot \left(1.714* 10^(-9)\,(BTU)/(h\cdot ft^(2)\cdot R^(4)) \right)\cdot [(554.07\,R)^(4)-(527.67\,R)^(4)]`

`\dot Q = 417.492\,(BTU)/(h)`

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:

`Q = \dot Q \cdot \Delta t` (2)

Where
`\Delta t` is the heat transfer time, measured in hours.

If we know that
`\dot Q = 417.492\,(BTU)/(h)` and
`\Delta t = (1)/(3)\,h`, then the net energy transfer is:

`Q = \left(417.492\,(BTU)/(h) \right)\cdot \left((1)/(3)\,h \right)`

`Q = 139.164\,BTU`

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.